Proof. If $n$ is a Carmichael number, then
$$b^{n-1} \equiv 1(\bmod n)$$
for all positive integers $b$ with $(b, n)=1$. Theorem 8.20 tells us that there is an integer $a$ with $\operatorname{ord}_{n} a=\lambda(n)$, where $\lambda(n)$ is the minimal universal exponent, and since $a^{n-1} \equiv 1(\bmod n)$, Theorem 8.1 tells us that
$$\lambda(n) \mid(n-1)$$

Now $n$ must be odd, for if $n$ was even, then $n-1$ would be odd, but $\lambda(n)$ is even (since $n>2$ ), contradicting the fact that $\lambda(n) \mid(n-1)$.

We now show that $n$ must be the product of distinct primes. Suppose $n$ has a prime-power factor $p^{t}$ with $t \geqslant 2$. Then
$$\lambda\left(p^{t}\right)=\phi\left(p^{t}\right)=p^{t-1}(p-1) \mid \lambda(n)=n-1$$

This implies that $p \mid(n-1)$, which is impossible since $p \mid n$. Consequently, $n$ must be the product of distinct odd primes, say
$$n=q_{1} q_{2} \cdots q_{k}$$

We conclude the proof by noting that
$$\lambda\left(q_{i}\right)=\phi\left(q_{i}\right)=\left(q_{j}-1\right) \mid \lambda(n)=n-1$$