Given that $p$ is a prime and $p\mid a^n$, prove that $p^n\mid a^n$. 
Given that $p$ is a prime and $p\mid a^n$, prove that $p^n\mid a^n$.

I know that the fundamental theorem of arithmetic states that any positive integer can be represented as a product of primes but how do I apply this to the proof? 
 A: Prime factorization and the Fundamental Theorem of Arithmetic can be used, but seem too high powered for this.
We can inductively show that if $p\mid a^n$ then $p\mid a$:

Clearly, it is true for $n=1$.


Suppose it is true for $n-1$, and $p\mid a^n=a\,a^{n-1}$. Since $p$ is prime, $p\mid a$ or $p\mid a^{n-1}$ either of which means that $p\mid a$.

Since $p\mid a^n$, we have that $p\mid a$ and therefore, $p^n\mid a^n$ (i.e. $a=kp\implies a^n=k^np^n$).
A: Use the fundamental theorem of arithmetic to uniquely express $a$ as a product of powers of primes
$$
a=p_1^{r_1}\dotsb p_\ell^{r_\ell}
$$
where $p_1<\dotsb<p_\ell$ and $r_i\geq1$ for $1\leq i\leq \ell$. The assumption $p\mid a^n$ then implies $p\mid p_k^{n r_k}$ for some $1\leq k\leq \ell$. It follows that $p=p_k$ since otherwise $p$ would not be prime. Hence $p^n=p_k^n$ which clearly divides $a^n$.
A: Since $p$ is prime and $p\mid a^n$, we have $p\mid a$. (Do you see why?) Thus $a=pk$ for some $k\in\mathbb{N}$, so $a^n=p^nk^n$. Hence $p^n\mid a^n$.
A: You don't need unique factorization, you just need Euclid's Proposition 12 from Book IX, that if $p\mid a^n$, then $p\mid a$, from which you can say that $a=pb$ for some $b$, hence $a^n=(pb)^n=p^nb^n$, which implies $p^n\mid a^n$, all by definition of divisibility.
