$\sum_{k=1} ^n (k^2 +1)k!=n(n+1)!$ I'm to prove this by mathematical induction:
Edited: I made a typo error. 
$\sum_{k=1} ^n (k^2+1)k!=n(n+1)!$
I made the test and the rightside is true. 
So I tested: $N+1$
$N(N+1)! + (N^2+1)N!$
$N(N+1)(N!)+(N^2+1)N!$
$N!\left[N(N+1)+N^2+1\right]$
$N! (N^2+N+N^2+1)$
$\sum_{k=1} ^ {N+1} (k^2+1)k!=$ $N!(2N^2+N+1)$
Is this a valid proof? 

$\sum _{k=1} ^{N+1} (k^2+1)k!$=$[(N+1)^2 +1]!+N(N+1)!\\
=(N+1)![(N^2+3N+2]\\
=(N+1)(N+1)(N+2)\\
=(N+1)(N+2)!$
