Proof. Let $b$ be a positive integer with $(b, n)=1$. Then $\left(b, q_{j}\right)=1$ for $j=1,2, \ldots, k$, and hence, by Fermat's little theorem, $b^{q_{j}-1} \equiv 1\left(\bmod q_{j}\right)$ for $j=1,2, \ldots, k$. Since $\left(q_{j}-1\right) \mid(n-1)$ for each integer $j=1,2, \ldots, k$, there are integers $t_{j}$ with $t_{j}\left(q_{j}-1\right)=n-1$. Hence, for each $j$, we know that $b^{n-1}=b^{\left(q_{j}-1\right) t_{j}} \equiv 1\left(\bmod q_{j}\right)$. Therefore, by Corollary 3.2 , we see that $b^{n-1} \equiv 1(\bmod n)$, and we conclude that $n$ is a Carmichael number.