At the heart this is really trivial, a consequence of $\rm\ f(x) = x^n\, \Rightarrow\ f{\:'}(x)\ =\ n\, x^{n-1}\ \equiv\ 0\:\ (mod\ n).\, $ First I give the simple proof, then I explain the viewpoint in terms of derivatives.
$\rm\ \ n\:|\:(a-b)\frac{a^n-b^n}{a-b}\ \Rightarrow\,\ n = m\:k,\ \ m\:|\:a-b,\ \ k\:\big|\frac{a^n-b^n}{a-b}\ $ by unique factorization. Thus it suffices
to show $\rm\ \ m\:|\:n,\:a-b\ \Rightarrow\ m\ \,\Big|\frac{a^n-b^n}{a-b}\: =\ a^{n-1}+a^{n-2}\:b+\:\cdots\:+a\:b^{n-2}+b^{n-1}$
But, $\rm\,\ mod\ \:m:\:\, \ a\equiv b\ \ \ \Rightarrow\, \ \ \frac{a^n-b^n}{a-b}\ \equiv\ a^{n-1}+\cdots+a^{n-1} \equiv\, n\, a^{n-1}\equiv\, 0\ $ via $\rm\ m\:|\:n\quad\ $ QED
The prior line is the special case $\rm\ f = x^n,\ x = b\ $ of this polynomial Taylor series approximant:
$\rm\displaystyle\quad\quad\quad\quad\quad \frac{f(x)-f(a)}{x-a} \: \equiv\ f\:'(a)\ \ \ (mod\ \:x-a)\quad$ for $\rm\ f(x)\in \mathbb Z[x]$
$\qquad $ i.e. $\rm\quad\ f(x)\ =\ f(a) +\: f\:'(a)\ (x-a) \:+\: (x-a)^2\: g(x)\quad$ for some $\rm\ g(x) \in \mathbb Z[x]$
Therefore this result about numbers is just a special case of the following well-known result about functions (here polynomials): $ $ a root $\rm\ x = a\ $ of $\rm\ f(x)\ $ has multiplicity $\rm > 1\ \iff\ f\:'(a) \:=\: 0.\:$ In fact, like above, many results about numbers are actually specializations of results about functions. Moreover, because functions have richer structure than numbers - e.g. having derivatives available - we can exploit this structure in the function realm before specializing to numbers. A powerful example of this is Mason's ABC theorem - which has a trivial high-school level proof for polynomials, but is an unproven conjecture for numbers. It yields, as a consequence, a trivial proof of FLT for polynomials. The moral is: to prove a result about numbers, try to interpret it as special case of a result about functions.