Yet another framing: The Gamma function is the only choice that satisfies the functional equation and "stays approximately exponential for large values", or "doesn't have oscillatory behavior for large values". I think the "log-convex" story is essentially a precise and quick shorthand for this, but it also obscures the motivation somewhat, and it's instructive to unpack it.

Roughly, in a relatively small interval around a large value on $n$, the behavior of $n!$ is roughly exponential. if $n$ is much larger than $k$, you can get a decent approximation of $(n+k)!$ by starting with n! and multiplying it by $n$, $k$ times.

More precisely, for a fixed integer $k$, as $n \to \infty$,

$$(n+k)! \approx n! \cdot n^k.$$

Even more precisely, for a fixed integer $k$, as $n \to \infty$

$$ \frac{(n+k)!}{n! \cdot n^k} \to 1.$$

If you're going to extend the factorial function to all positive real numbers, you might hope to preserve this property for non-integer values of $k$ and $n$.

Put a slightly different way, for large $n$, we have:

• $(n+1)!$ is roughly halfway between $n!$ and $(n+2)!$ multiplicatively
• $(n+30)!$ is roughly $3/10$ of the way from $n!$ to $(n+100)!$ multiplicatively.

You might sensibly demand:

• $(n+1/2)!$ is roughly halfway between $n!$ and $(n+1)!$ multiplicatively.

So you might require that for large $n$, we have $(n+1/2)! \approx n! \cdot n^{1/2}$, i.e. as $n \to \infty$

$$\frac{(n+1/2)!}{n! \cdot n^{1/2}} \to 1.$$

Combined with the functional equation $t! = (t-1)! \cdot t,$ this forces you you to have

$$\frac{(1/2)! \cdot (3/2) \cdot (5/2) \cdot \dots \cdot (n-1/2) \cdot (n+1/2)}{\left(1\cdot 2 \cdot \dots \cdot n \right) \cdot n^{1/2}} \to 1,$$

i.e.

$$(1/2)! = \lim_{n\to\infty} \frac{\left(1\cdot 2 \cdot \dots \cdot n \right) \cdot n^{1/2}}{ (3/2) \cdot (5/2) \cdot \dots \cdot (n-1/2) \cdot (n+1/2)}.$$

So picking a sensible behavior for factorial at large inputs and then pulling that information "back" to the value at 1/2 using the functional equation forces you to set $(1/2)!$ to whatever the value of that limit is, which happens to be $\sqrt{pi}/2$.

If you picked any other value for $(1/2)!$, it would get "pushed" to large values of $n$ in an upsetting way where taking a half-integer step from $n!$ to $(n+1/2)!$ makes a multiplicative step that's larger or smaller by some fixed factor than taking the next half-integer step from $(n+1/2)!$ to $(n+1)!$, contrary to the nice regular behavior for integer steps.

Yet another framing: The Gamma function is the only choice that satisfies the functional equation and "stays approximately exponential for large values", or "doesn't have oscillatory behavior for large values". I think the "log-convex" story is essentially a precise and quick shorthand for this, but it also obscures the motivation somewhat, and it's instructive to unpack it.

Roughly, in a relatively small interval around a large value on $n$, the behavior of $n!$ is roughly exponential. if $n$ is much larger than $k$, you can get a decent approximation of $(n+k)!$ by starting with n! and multiplying it by $n$, $k$ times.

More precisely, for a fixed integer $k$, as $n \to \infty$,

$$(n+k)! \approx n! \cdot n^k.$$

Even more precisely, for a fixed integer $k$, as $n \to \infty$

$$ \frac{(n+k)!}{n! \cdot n^k} \to 1.$$

If you're going to extend the factorial function to all positive real numbers, you might hope to preserve this property for non-integer values of $k$ and $n$.

Put a slightly different way, for large $n$, we have:

• $(n+1)!$ is roughly halfway between $n!$ and $(n+2)!$ multiplicatively
• $(n+30)!$ is roughly $3/10$ of the way from $n!$ to $(n+100)!$ multiplicatively.

You might sensibly demand:

• $(n+1/2)!$ is roughly halfway between $n!$ and $(n+1)!$ multiplicatively.

So you might require that for large $n$, we have $(n+1/2)! \approx n! \cdot n^{1/2}$, i.e. as $n \to \infty$

$$\frac{(n+1/2)!}{n! \cdot n^{1/2}} \to 1.$$

Combined with the functional equation $t! = (t-1)! \cdot t,$ this forces you you to have

$$\frac{(1/2)! \cdot (3/2) \cdot (5/2) \cdot \dots \cdot (n-1/2) \cdot (n+1/2)}{\left(1\cdot 2 \cdot \dots \cdot n \right) \cdot n^{1/2}} \to 1,$$

i.e.

$$(1/2)! = \lim_{n\to\infty} \frac{\left(1\cdot 2 \cdot \dots \cdot n \right) \cdot n^{1/2}}{ (3/2) \cdot (5/2) \cdot \dots \cdot (n-1/2) \cdot (n+1/2)}.$$

So picking a sensible behavior for factorial at large inputs and then pulling that information "back" to the value at 1/2 using the functional equation forces you to set $(1/2)!$ to whatever the value of that limit is, which happens to be $\sqrt{\pi}/2$.

If you picked any other value for $(1/2)!$, it would get "pushed" to large values of $n$ in an upsetting way where taking a half-integer step from $n!$ to $(n+1/2)!$ makes a multiplicative step that's larger or smaller by some fixed factor than taking the next half-integer step from $(n+1/2)!$ to $(n+1)!$, contrary to the nice regular behavior for integer steps.