Mathematical induction last step...

Mathematical induction last step...

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?
$$1(1!)+2(2!)+3(3!)+...+k(k!)+(k+1)(k+1)!=(k+1)!(k+2)-1)$$