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.
$\blacksquare$