Number of permutations increases $56$ times when number of elements increases by $2$, how many elements?

Question

I have to find the number of elements $n$ so that when their number is increased by $2$, the number of their permutations $m$ increases $56$ times. Can I write it down with the following equation?

$(n+2)! = 56m$

If so, how can I simplify it to compute the number of elements n?

Answer 1

$(n + 2)! = 56 n!$

$(n + 2) (n + 1) n! = 56 n!$

$(n + 2) (n + 1) = 56$

Solving for $n \in \mathbb{N}$ yields $\boxed{n = 6}$.