Is it a theorem that $\ n^{(n-1)^p} \equiv 1\pmod {n-1} $

Question

Is it a theorem that $\ n^{(n-1)^p} \equiv 1\pmod {n-1} $ $\ n,p \in \mathbb Z^+ $

For example $$\ 4^{3^p} \equiv 1\pmod {3} $$

This seems to be true empirically but I have no idea if it has been proven. I can't find anything about it. I was also wondering if $\ a = n, \ b =n-1 \ $are the only solutions to the congruence$$\ a^b \equiv 1\pmod b$$ Any information is appreciated. I now realize that my first question is obvious but am really more interested in my second question about the positive integer solutions of the above congruence.

Answer 1

If $a \equiv 1 \mod b$ then $a^k \equiv 1 \mod b$ for all $k>1$. This is because $\mathbb Z / b \mathbb Z$ (the integers "modulo b") forms a ring. We can prove that statement quite easily for $k\geq 0$: Since $a \equiv 1 \mod b$ there is some $m$ such that $a + m\cdot b=1$. Therefore

$$a^k = (a+mb)^k = \sum_{n=0}^k \binom{k}{n} \underbrace{a^n (mb)^{k-n}}_{\text{divisible by $b$ for all $n<k$}} $$

and so $a^k \equiv a \equiv 1 \mod b$.

To answer your second question: No that is not necessarily the only solution. Consider the case $3^4 \equiv 81 \equiv 1 \mod 4$.