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? 

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. $a=kp\implies a^n=k^np^n$).
