We could use a combinatorial argument to prove that:
$n!(n+1) = (n+1)!$
This may not be what the PO wants, nevertheless, I found it interesting!
is to construct 2 scenarios for generating permutation.
Scenario 1 takes (n+1) objects to permute. The number of permutation is $(n+1)!$ by definition. This covers the right hand side of the equation (see fig.1)
Scenario 2 permutes the (n+1) objects in 2 steps. The first step permutes $n$ objects and in the second step tries to create the final permutation using $(n+1)$ objects. As an example, let $n+1=3$, so $n=2$, and we want to generate permutations for A,B,C (n+1 objects). To do that we'll create 2 sets from {A,B,C}. Namely {A,B} and {C}.
{A,B} can be permuted in $2!=2*1=2$ different ways as in Fig.2, but we are after creating all possible combinations form {A,B,C}, so we need to use C with each of the generated permutations {A,B} and {B,A}.
We can merge C in the permutation A,B for example, in 3 ways (that is n+1 ways), to obtain 3 new permutations,namely: (C, A, B), (A, C, B) and (A, B, C). This merging process can repeat for each row of the $n!$ rows resulting from permuting the $n$ objects.
That is, we end up with $n! *(n+1)$ permutations, which is the left hand side of the equation.
