This is another answer in terms of Poisson processes and the Gamma distribution, and it still uses a bit of calculus which you might call a trick, but I think at least it does add another bit of intuition:

Consider the homogeneous Poisson process with rate parameter 1; this means we are counting the number of occurrences of an event that happens with rate 1. Let's calculate the probabilities $p_k(t)$ that we are in state $k$ at time $t$, i.e., that the event occurs $k$ times in the interval $[0,t]$.

Since the event happens with rate 1, probability mass flows from $p_k(t)$ to $p_{k+1}(t)$ with rate 1. This means that ${p'}_0(t) = -p_0(t)$ and ${p'}_{k+1}(t) = p_k(t) - p_{k+1}(t)$. Also, $p_0(0) = 1$ and $p_{k+1}(0) = 0$. Here comes the bit of calculus: these equations have the solution $p_k(t) = \frac{t^k}{k!} e^{-t}$. (A bit fuzzily, we can read this in two parts: $\frac{t^k}{k!}$ is 1 integrated $k$ times, and $e^{-t}$ represents probability mass being lost at rate 1 to states further down the line. See below for yet another fuzzy explanation.)

Now consider the waiting time $T_k$ until the $k$'th occurrence. Clearly, $T_k = t$ means that the transition from state $k-1$ to state $k$ happens at time $t$, so the probability of $T_k \le t$ is the probability that the transition happens before time $t$, and the density is the derivative of this, i.e., the rate at which probability mass flows from state $k-1$ to $k$. This equals the occurrence rate (i.e., 1) times $p_{k-1}(t)$.

So the probability density of the random variable $T_k$ is $1 \cdot p_{k-1}(t) = \frac{t^{k-1}}{(k-1)!} e^{-t}$ (for $t \ge 0$). Since the probability that there is no occurrence ever is obviously zero, $\int_{0}^{\infty} \frac{t^{k-1}}{(k-1)!} e^{-t} = 1$.

Incidentally, this is related to a way of thinking about why $\sum_{k=0}^\infty \frac{t^k}{k!} = \lim_{n\to\infty} (1 + \frac{t}{n})^n$. Suppose you start with one unit of money in an account and get 100% interest, continuously compounded. However, the interest from the original account (number 0) is paid not the original account, but to account #1; interest from account #1 is paid to account #2, and so on.

Then the money $m_k(t)$ in account $k$ equals $1$ integrated $k$ times, and the total money $m(t)$ is $\sum_{k=0}^\infty m_k(t) = \sum_{k=0}^\infty \frac{t^k}{k!}$. But on the other hand, $m(t)$ is continuously compounded at 100% interest, so $m(t) = \lim_{n\to\infty} (1 + \frac{t}{n})^n$ by the usual reasoning.

This gives another fuzzy argument why we should have $p_k(t) = \frac{t^k}{k!} e^{-t}$. The change in the $p_k(t)$ over time consists of two parts: on the other hand, each $p_{k+1}(t)$ increases at rate $p_k(t)$; on the other hand, each $p_k(t)$ decreases at rate $p_k(t)$. If we view the $p_k(t)$ as accounts, since we are taking money out of every account at the constant rate 1, the effect is to decrease the total amount at the constant rate 1, i.e. by a factor of $e^{-t}$, which cancels out the increase of $e^t$ due to the accruing interest. It makes some intuitive sense that we can model this effect by simply rescaling the amount of money in each of the accounts by $e^{-t}$.

This is another answer in terms of Poisson processes and the Gamma distribution, and it still uses a bit of calculus which you might call a trick, but I think at least it does add another bit of intuition:

Consider the homogeneous Poisson process with rate parameter 1; this means we are counting the number of occurrences of an event that happens with rate 1. Let's calculate the probabilities $p_k(t)$ that we are in state $k$ at time $t$, i.e., that the event occurs $k$ times in the interval $[0,t]$.

Since the event happens with rate 1, probability mass flows from $p_k(t)$ to $p_{k+1}(t)$ with rate 1. This means that ${p'}_0(t) = -p_0(t)$ and ${p'}_{k+1}(t) = p_k(t) - p_{k+1}(t)$. Also, $p_0(0) = 1$ and $p_{k+1}(0) = 0$. Here comes the bit of calculus: these equations have the solution $p_k(t) = \frac{t^k}{k!} e^{-t}$. (A bit fuzzily, we can read this in two parts: $\frac{t^k}{k!}$ is 1 integrated $k$ times, and $e^{-t}$ represents probability mass being lost at rate 1 to states further down the line. See below for yet another fuzzy explanation.)

Now consider the waiting time $T_k$ until the $k$'th occurrence. Clearly, $T_k = t$ means that the transition from state $k-1$ to state $k$ happens at time $t$, so the probability of $T_k \le t$ is the probability that the transition happens before time $t$, and the density is the derivative of this, i.e., the rate at which probability mass flows from state $k-1$ to $k$. This equals the occurrence rate (i.e., 1) times $p_{k-1}(t)$.

So the probability density of the random variable $T_k$ is $1 \cdot p_{k-1}(t) = \frac{t^{k-1}}{(k-1)!} e^{-t}$ (for $t \ge 0$). Since the probability that there is no occurrence ever is obviously zero, $\int_{0}^{\infty} \frac{t^{k-1}}{(k-1)!} e^{-t} dt = 1$.

Incidentally, this is related to a way of thinking about why $\sum_{k=0}^\infty \frac{t^k}{k!} = \lim_{n\to\infty} (1 + \frac{t}{n})^n$. Suppose you start with one unit of money in an account and get 100% interest, continuously compounded. However, the interest from the original account (number 0) is paid not the original account, but to account #1; interest from account #1 is paid to account #2, and so on.

Then the money $m_k(t)$ in account $k$ equals $1$ integrated $k$ times, and the total money $m(t)$ is $\sum_{k=0}^\infty m_k(t) = \sum_{k=0}^\infty \frac{t^k}{k!}$. But on the other hand, $m(t)$ is continuously compounded at 100% interest, so $m(t) = \lim_{n\to\infty} (1 + \frac{t}{n})^n$ by the usual reasoning.

This gives another fuzzy argument why we should have $p_k(t) = \frac{t^k}{k!} e^{-t}$. The change in the $p_k(t)$ over time consists of two parts: on the other hand, each $p_{k+1}(t)$ increases at rate $p_k(t)$; on the other hand, each $p_k(t)$ decreases at rate $p_k(t)$. If we view the $p_k(t)$ as accounts, since we are taking money out of every account at the constant rate 1, the effect is to decrease the total amount at the constant rate 1, i.e. by a factor of $e^{-t}$, which cancels out the increase of $e^t$ due to the accruing interest. It makes some intuitive sense that we can model this effect by simply rescaling the amount of money in each of the accounts by $e^{-t}$.