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Fermat's little theorem states that, for a prime number $p$ and any integer $a$,If $p\mid a$, both sides vanish modulo $p$. Otherwise multiplication by $a$ permutes the nonzero residue classes modulo $p$: it is injective because $ai\equiv aj$ implies $i\equiv j$ by cancellation. Taking the products of these residues givesThe factorial is coprime to $p$, so cancellation proves $a^{p-1}\equiv 1(\mathrm{mod}p)$ and hence the stated version. This proves Fermat's little theorem in both cases.

Let $n$ be a Carmichael number. If a prime square $p^2$ divided $n$, choose $a=p$. Since $n$ is an odd composite integer, $n>1$, and $p^n-p\equiv -p\not\equiv 0(\mathrm{mod}{p}^2)$. This contradicts the required modular congruence modulo $n$. Therefore every Carmichael number is squarefree.

For each prime divisor $p$ of $n$, choose the allowed primitive root $g$ modulo $p$. The Carmichael identity applied to the positive integer $g$ gives $g^n\equiv g(\mathrm{mod}p)$, and cancellation gives $g^{n-1}\equiv 1(\mathrm{mod}p)$. By its defining order property,Conversely, suppose a composite integer $n$ is squarefree and has this divisibility for every prime divisor. For an arbitrary integer $a$, modulo a prime divisor $p$ there are two cases. If $p\mid a$, then $a^n\equiv a\equiv 0$. Otherwise Fermat's little theorem and $p-1\mid n-1$ give $a^{n-1}\equiv 1$, so again $a^n\equiv a$. Since the prime divisors are distinct and pairwise coprime, their product $n$ divides $a^n-a$. This is the proof of Korselt criterion:

If $n=pq$ for distinct odd primes with $p<q$, the condition at $q$ would require $q-1\mid pq-1$. Reducing $pq-1$ modulo $q-1$ gives $p-1$, so $q-1\mid p-1$. But $0<p-1<q-1$, which is impossible. Thus a product of two distinct odd primes is not Carmichael.

For $n=pqr$, squarefreeness is automatic. Since $p\equiv 1(\mathrm{mod}p-1)$,Apply the analogous reductions at $q$ and $r$ to obtainFinally $1729=7\cdot 13\cdot 19$, with distinct odd prime factors, andAll three divisibilities hold, so 1729 is a Carmichael number.