TITLE:  How to prove: $n! \times (n+1) = (n+1)!$
QUESTION [7 upvotes]: Plugging in numbers shows me that it is the same, but I don't know how to get from one to the other.
I get stuck at:
$$
n!(n+1) = n n! + n!
$$
I can't make the connection for the final step.
$(2n + n)!$ doesn't seem right, cause then i would just take n out again and be left with  $n!$ $(n+1)$ which is the same thing that I started with...yeah for U turns.

REPLY [7 votes]: Yet another way to write out a "proof" would be using product notation and the definition $n! = \prod_{k=1}^n k$:
$$(n+1)\,n! = (n+1) \prod_{k=1}^n k = \prod_{k=1}^{n+1} k = (n+1)!$$
But of course, as Chris Taylor points out in the comments, the validity of proofs like this really depends on what definition of the factorial you're starting from, and on what lemmas you may take as already proven.