number theory - Show that $a^n-b^n$ has a prime factor which does not divide $a-b$ for all $n>1$ .

I was asked to prove the following using the lifting the exponent lemma.

Show that $a^n-b^n$ has a prime factor which does not divide $a-b$ for all $n>1$ .

Using the first lemma, what I got was this:
if $p$ is any prime greater than $2$,

then we have

$V_p(a^n-b^n)= V_p(a-b) + V_p(n)$

where $V_p(x)$ is the highest power of $p$ that divides $x$ and $p|a-b$ but does not divide a or b.
I don't know how to approach this and would welcome some hints.

Answer

Hint:

$$\frac {a^n-b^n}{a-b}=\sum_{k=1}^n a^{k-1}b^{n-k}>n\ge\prod_{p|(a-b)} p^{V_p(n)}.$$

$$\implies a^n-b^n>\prod_{p|(a-b)} p^{V_p(a^n-b^n)}$$