I have a quick (but messy) intuition for this property(?), which goes as follows.
Given $a^n$, then $a^n = a \cdot a \, \cdot \, ... \, \cdot \, a$, where there are $n - 1$ multiplication operations. $a$ has prime factors $x_1, x_2, ..., x_i$, where each factor has its own power $m$, that is, $a = x_1^{m_1} \cdot x_2^{m_2} \, \cdot \, ... \, \cdot \, x_i^{m_i}$.
With that, we have: \begin{align} a^n &= a \cdot a \, \cdot \! \underbrace{...}_\text{n times} \! \cdot \, a \\ &= (x_1^{m_1} \cdot x_2^{m_2} \, \cdot \, ... \, \cdot \, x_i^{m_i}) \cdot (x_1^{m_1} \cdot x_2^{m_2} \, \cdot \, ... \, \cdot \, x_i^{m_i}) \cdot \! \underbrace{...}_\text{n times} \! \cdot \, (x_1^{m_1} \cdot x_2^{m_2} \, \cdot \, ... \, \cdot \, x_i^{m_i}) \\ &= (x_1^{m_1} \cdot x_2^{m_2} \, \cdot \, ... \, \cdot \, x_i^{m_i})^n \\ &= x_1^{m_1n} \cdot x_2^{m_2n} \, \cdot \, ... \, \cdot \, x_i^{m_in} \end{align}
Here, we can see that every prime factor of $a^n$, $x_k$ for some $k$, is raised to a power that has factor of $n$. Therefore, $n$ divides the multiplicity of all factors of $a^n$.
Would this be enough for a proof?
Your proof is nearly good! Here is a cleaner version.
Let $p$ be a prime dividing $a$, where $p$ has multiplicity $k$ as a factor of $a$. This means $p^k | a$ but $p^{k+1} \nmid a$. Then $p^{nk} \mid a^n$ and $p^{nk + 1} \nmid a^n$ (if this last claim is not obvious to you, you should prove it. It isn't too hard). That is, $p$ has multiplicity $nk$ in the factorisation of $a^n$. This holds for any of the prime factors of $a$, so the result is proved.