If $a \mid b$, then $n^a \mid n^b$. Let $n, a, b$ be integers such that $a \mid b$. Does it follow that $n^a \mid n^b$ for any integer $n$?
I believe it does and here is what I have tried. From $ a \mid b$ we have that $a \mid a-b$. Now, $\frac{n^a} {n^b} =n^{a-b} $ and here I am stuck. 

If $a\mid b$ holds and $b\ne 0$ , then there is a positive integer $k$ with $b=ak$ , hence $$n^b=n^{ak}=(n^a)^k$$ implying $n^a\mid n^b$
