TITLE: Proof by induction, 1 · 1! + 2 · 2! + ... + n · n! = (n + 1)! − 1
QUESTION [9 upvotes]: So I'm supposed to prove that $$1 · 1! + 2 · 2! + \dots + n · n! = (n + 1)! − 1$$ using induction. 
What I've done
Basic Step:
Let $n=1$, $$1\cdot1! = 1\cdot1 = 1 = (n+1)!-1 = 2!-1 = 2-1  = 1$$
Induction Step:
Assume $f(k) = 1\cdot1! + 2\cdot2! + \dots + k\cdot k! = (k+1)!-1$
\begin{align}
F(k+1) &= 1\cdot1! + 2\cdot2! + \dots + k\cdot k! + (k+1)\cdot(k+1)!\\
&= (k+1)!\ - 1 + (k+1)\cdot(k+1)!\\
&= (k+1)!\cdot((k+1) - 1) = (k+1)!\cdot(k)
\end{align}
I think I'm supposed to make $(k+1)!\cdot k = ((k+1)+1)!+1 = (k+2)!-1$ but I'm not sure how to get there.

REPLY [5 votes]: Bernard's answer highlights the key algebraic step, but I thought I might mention something that I have found useful when dealing with induction problems: whenever you have an induction problem like this that involves a sum, rewrite the sum using $\Sigma$-notation. It makes everything more concise and easier to manipulate:
\begin{align}
\sum_{i=1}^{k+1}i\cdot i!&=\sum_{i=1}^k i\cdot i!+(k+1)(k+1)!\tag{by definition}\\[1em]
&= [(k+1)!-1]+(k+1)(k+1)!\tag{induction hyp.}\\[1em]
&= (k+1)![1+(k+1)]-1\tag{rearrange}\\[1em]
&= (k+1)![k+2]-1\tag{simplify}\\[1em]
&= (k+2)!-1.\tag{by definition}
\end{align}