Divisors of Factorial

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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.

$\blacksquare$