Remark that we need \frac{a^n - 1}{a-1} \equiv 0 \pmod{n}. Then remark that v_p(a-1) \cdot k \ge m. If p is an odd prime we have v_p(a^n - 1) = v_p(a-1) + m by LTE. But then v_p \left ( \frac{a^n - 1}{a-1} \right ) = m so p^m divides it. Hence we have shown all odd primes work out in the divisibility.
Now suppose 2|n, so that a is odd. But by LTE v_2(a^n - 1) = v_2(a-1) + v_2(a+1) + v_2(n) - 1 so v_2 \left ( \frac{a^n - 1}{a-1} \right ) = v_2(a+1) + v_2(n) - 1 \ge v_2(n), so enough factors of 2 divide it. Hence the result follows.
Let p be a prime divisor of n. It suffices to prove that \nu_{p}(n)\le\nu_{p}(a^{n-1}+a^{n-2}+\dots+a+1)=\nu_{p}\left(\frac{a^n-1}{a-1}\right),which is true by Theorem 6.22 of Number Theory: Concepts and Problems since p\mid a-1.\square
Isn't this problem just blind LTE?
Let a prime p \mid n. Then p divides a-1, so \nu_p\left(\frac{a^n-1}{a-1}\right) = \nu_p(a^n - 1) - \nu_p(a-1) = \nu_p(n)by LTE, as required.