If $p$ is prime and $p$ divides $a^n$ is it true that $p^n$ divides $a^n$? I'm stuck on this problem. A hint is to use the following corollary: 
If $p$ is prime and $p$ divides $(a_1 \times a_2\times \dotsb\times a_n)$, then $p$ divides at least one of the $a_i$ .
So Assume $p$ is prime and $p$ divides $a^n$. Then, $p = k \times a^n$ for some integer $k$. Now, by the corollary, it is assumed that $p$ also divides on of the $a_i$, take $i = 1$. Then, $p = c \cdot a_1$. This is where I get stuck and keep going in circles. Any advice would be much appreciated. 
 A: 
Theorem.
Let $c,b\in \Bbb Z$. If $p|cb$ then $p|c$ or $p|b$ (This is just a reformulation of yours).

Now, by induction, we want to prove that if $p|a^n$, $p|a$.
Base case is trivial, with $n=0,1$.
Now, if $p|a^n=a\cdot a^{n-1}$ then we have that (1) $p|a$ or (2) that $p|a^{n-1}$.
If it's (1), we're done. If it's (2), by inductive hypothesis we have that $p|a$.
A: Yes. Consider prime factorization of $a$,
$a= p{_1}^{k_1} p{_2}^{k_2} p{_3}^{k_3}.... p{_r}^{k_r}$, then 
$a^n= p{_1}^{nk_1} p{_2}^{nk_2} p{_3}^{nk_3}.... p{_r}^{nk_r}$. 
So if $p$ divides $a^n$, then $p$ divides one of the $p{_i}^{nk_i}$. Since $p$ and $p_{i}$ both are primes, thus it is only possible when   $p = p_i$. Since $p{_i}^{nk_i}=p^{nk_i}$ divides $a^n$, hence $p^n$ must divide $a^n$
A: If $p|a$ then $a = p*c$ and $a^n = (p*c)^n = p^n*c^n$
Can you show that $p|a$?
A: Let $K(a)= \{ x \in \mathbf{N} : x \mid a, x \in \mathtt{PRIMES} \}$, then $K(a) = K(a^n)$.
We can see that $p \mid a$ if and only if $p \in K(a)$.
So $p \mid a^n \Leftrightarrow p \in K(a^n) \Leftrightarrow p \in K(a) \Leftrightarrow p \mid a \Rightarrow p^n \mid a^n$.
