Divisors of Factorial

Theorem
Let $n \in \N_{>0}$.

Then all natural numbers less than or equal to $n$ are divisors of $n!$:


 * $\forall k \in \left\{{1, 2, \ldots, n}\right\}: n! \equiv 0 \pmod k$

Proof
From the definition of factorial:


 * $n! = 1 \times 2 \times \cdots \times \left({n-1}\right) \times n$

Thus every number less than $n$ appears as a divisor of $n!$.

The result follows from definition of congruence.