Since each term divides the next, it suffices to prove that for every positive integer $k$, there exists a prime divisor of $n^{n^k} - 1$ that does not divide $n^{n^{k-1}} - 1$. Assume the contrary. By LTE, for any prime $p$ dividing $n^{n^{k-1}} - 1$, we have \[v_p(n^{n^k} - 1) = v_p(n^{n^{k-1}} - 1) + v_p(n).\]Since $gcd(n^{n^{k-1}} - 1, n) = 1$, then $p \nmid n$, so \[v_p(n^{n^k} - 1) = v_p(n^{n^{k-1}} - 1).\]But since $n^{n^{k-1}} - 1$ and $n^{n^k} - 1$ have the same set of distinct prime divisors, we must have $n^{n^{k-1}} - 1 = n^{n^k} - 1$, which is a clear contradiction. Hence, we are done.

Zigmondy