Mathematical induction proof that $\forall{n}\in{N}:1(1!)+2(2!)+3(3!)+...+n(n!)=(n+1)!-1$ Proof:
$$\forall{n}\in{N}:1(1!)+2(2!)+3(3!)+...+n(n!)=(n+1)!-1$$
Step 1
Prove p(n)=>True for n=1
$$1(1!)=(1+1)!-1$$
$$1=1$$
Step 2
Assume by induction that (k)=>true
$$\forall{k}\in{N}:1(1!)+2(2!)+3(3!)+...+k(k!)=(k+1)!-1$$
Step 3
Prove p(k+1)=>true
$$1(1!)+2(2!)+3(3!)+...+k(k!)=(k+1)!-1$$
$$1(1!)+2(2!)+3(3!)+...+k(k!)+(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)!$$
Question
Is this the end of the demonstration?
$$1(1!)+2(2!)+3(3!)+...+k(k!)+(k+1)(k+1)!=(k+1)!(k+2)-1)$$

The last term should be $(k + 1)! \cdot (k+2) - 1 = (k + 2)! - 1$ instead of $(k+1)! \cdot (k+1) - 1$, but aside from that it is just fine.
