15. Proof: From $a^{m-1} \equiv 1(\bmod m)$, we know that $(a, m)=1$. By Theorem 1 of Chapter 5, we have
$$a^{\varphi(m)} \equiv 1(\bmod m)$$

Let $d$ be the smallest positive integer solution to the congruence $a^{x} \equiv 1(\bmod m)$, then it must be that $d \mid \varphi(m)$, because otherwise it can be written as $\varphi(m)=d q+r, 01$, then $p_{i}<m$, $1 \leqslant i \leqslant n$. Therefore,
$$\begin{aligned} \varphi(m)= & m\left(1-\frac{1}{p_{1}}\right)\left(1-\frac{1}{p_{2}}\right) \cdots\left(1-\frac{1}{p_{n}}\right) \\ & <m\left(1-\frac{1}{m}\right)^{n}<m\left(1-\frac{1}{m}\right)=m-1 \end{aligned}$$

This contradicts $(m-1) \mid \varphi(m)$. Therefore, $m$ must be a prime number.