I don't think your proof is formal enough. Let us examine it:
Let $m \in \mathbb{N}^*$ and $m = p^{e_{1}}_{1} \cdot p^{e_{2}}_{2} \cdot ... \cdot p^{e_{k}}_{k}$ by the fundamental theorem of arithmetic.
You're given that $m = p^{e_{1}}_{1} \cdot p^{e_{2}}_{2} \cdot ... \cdot p^{e_{k}}_{k}$ in the exercise; However, if you still want to mention the fundamental theorem, I prefer it to be: "let $m \in \mathbb{N}^*$; by the fundamental theorem of arithmetic, $m$ has a unique factorization $m = p^{e_{1}}_{1} \cdot p^{e_{2}}_{2} \cdot ... \cdot p^{e_{k}}_{k}$".
We know that $p$ divides $p^{e_{1}}_{1} \cdot p^{e_{2}}_{2} \cdot ... \cdot p^{e_{k}}_{k}$ therefore $p$ divides $m$.
It may be just "It is given that $p$ divides $m = p^{e_{1}}_{1} \cdot p^{e_{2}}_{2} \cdot ... \cdot p^{e_{k}}_{k}$".
Because of the added fact $p$ is prime that means that $p$ is included in the sequence that forms m by Euclid's theorem. So we can see that $p$ divide $p^{e_{1}}_{1}$ or $p^{e_{2}}_{2}$ or... so one. Because $m$ is formed of prime numbers multiplied together it is necessary that it will contain $p$ in it if $p$ divide $m$. So we can conclude that here exists $1 \leq i \leq k$ such that $p = p_i$ is true.
Here is the considerable problem in your proof; You haven't shown that explicitly. You should write: "$p$ is prime and it divides $m$, so there exists some $i$ such that $p \vert p_i$ - we will show a stronger statement: If $m=m_1 \cdot ... \cdot m_t$ for some integers $m_1, ... m_t$ (not necessarily primes), then $p \vert m_i$ for some $i$. We show this by induction on $k$:
If $k=1$, that is $m=m_1$, then $p \vert m_1 = m$.
Let $k$ be an arbitrary integer, $m = m_1 \cdot ... \cdot m_k$; Let us divide $m$ by $p$: $m_1 \cdot ... \cdot m_k = p \cdot x$ for some $x$. If $p$ divides $m_1$ we are done, otherwise, as $p$ divides a multiplication of two numbers $m_1 \cdot (m_2 \cdot ... \cdot m_k)$, and by Euclid's lemma $p$ divides one of them (or both); But as $p$ doesn't divide $m_1$, it divides $(m_2 \cdot ... \cdot m_k)$, and finally by inductive hypothesis, there exists some $2 \leq i \leq k$ such that $p \vert m_i$.
It finishes the proof because $m = p_1 \cdot ... \cdot p_1 \cdot ... \cdot p_k \cdot ... \cdot p_k$ and we have shown that $p \vert p_i$ for some $i$. The only two divisors of a prime number ($p_i$) are $1$ and $p_i$, but $p$ is prime and therefore cannot be $1$. So, $p = p_i$".
If you are not allowed to use Euclid's lemma (that says that if a prime $p$ divides $a \cdot b$ then $p \vert a$ or $p \vert b$), prove it.
