$\sum_{k=1} ^n (k^2 +1)k!=n(n+1)!$ 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? 
 A: There are various problems, including 


*

*the thing you are trying to prove is not correct, for example when $n=2$

*you did not end up showing the inductive hypothesis was true for $N+1$

*you seem to have added $(N^2+1)N!$ rather than $(N+1+1)\,(N+1)!$
So let's illustrate proving something that is true: $$\sum\limits_{k=1} ^n k\cdot k!=(n+1)!-1$$
First we check that it is correct at the start, when $n=1$: we have $1 \cdot 1!=1$ on the left hand side and $2!-1=1$ on the right and these are indeed equal
Then we assume it is true for $n=m$ and consider what happens trying to go forward a step:
$\sum\limits_{k=1} ^{m+1} k\cdot k! = \sum\limits_{k=1} ^m k\cdot k! + (m+1)\cdot (m+1)! \\ =(m+1)!-1+(m+1)\cdot (m+1)! \\= (m+1)!(1+m+1)-1 \\= (m+2)! -1$
which shows it would also be true for  $n=m+1$ 
and therefore, by induction, it is true for all positive integer $n$ 
A: Your inductive step doesn't work because your expression doesn't simplify to $(N+2)!(N+1)$.
Note that $$\sum_{k=1}^n(k^2+1)k!=\sum_{k=1}^n((k+2)!-3(k+1)!+2\cdot k!)\\=\sum_{k=1}^n((k+2)!-(k+1)!+2\cdot k!-2\cdot (k+1)!)\\=(n+2)!-2\cdot (n+1)!=(n+1)!n.$$
A: $\sum _{k=1} ^{N+1} (k^2+1)k!$=$[(N+1)^2 +1]!+N(N+1)!\\
=(N+1)![(N^2+3N+2]\\
=(N+1)(N+1)(N+2)\\
=(N+1)(N+2)!$
