TITLE: Prove by Mathematical Induction: $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$
QUESTION [7 upvotes]: Prove by Mathematical Induction . . .
$1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$
I tried solving it, but I got stuck near the end . . .
a. Basis Step: 

$(1)(1!) = (1+1)!-1$
$1 = (2\cdot1)-1$
$1 = 1 \checkmark$

b. Inductive Hypothesis

$1(1!) + 2(2!) + \cdot \cdot \cdot +k(k!) = (k+1)!-1$

Prove k+1 is true.

$1(1!) + 2(2!) + \cdot \cdot \cdot +(k+1)(k+1)! = (k+2)!-1$
$\big[RHS\big]$
$(k+2)!-1 = (k+2)(k+1)k!-1$
$\big[LHS\big]$
$=\underbrace{1(1!) + 2(2!) + \cdot \cdot \cdot + k(k+1)!} + (k+1)(k+1)!$ (Explicit Last Step)
$= \underbrace{(k+1)!-1}+(k+1)(k+1)!$ (Inductive Hypothesis Substitution)
$= (k+1)!-1 + (k+1)(k+1)k!$
$= (k+1)k!-1 + (k+1)^{2}k!$

My [LHS] looks nothing like my [RHS] did I do something wrong?
EDIT:

$ = (k+1)k! + (k+1)^2k! -1 $
$ = (k+1)(k!)(1 + (k+1))-1$
$ = (k+1)(k!)(k+2)-1 = (k+2)(k+1)k!-1$

REPLY [5 votes]: It's a special case of telescopic induction. This post has a very short and simple inductive proof of
Theorem $\rm\displaystyle\,\ \sum_{i\,=\,1}^n f(i) = g(n)\iff g(1) = f(1)\ {\rm and}\ g(n\!+\!1)-g(n) = f(n\!+\!1)\ $ for $\,n \ge 1.$
Applied to your case, where $\rm\,f(n) = n n!\:$ and $\rm\:g(n) = (n+1)!-1,\,$ we have $\rm\: g(1)=1 = f(1),\:$ and 
$$\begin{eqnarray}\rm g(n\!+\!1)-g(n) &=&\rm (n\!+\!2)!-1-((n\!+\!1)!-1) \\
 &=&\rm (n\!+\!2)!\,-\,(n\!+\!1)! \\
 &=&\rm (n\!+\!2 -1)(n\!+\!1)! \\
 &=&\rm (n\!+\!1)(n\!+\!1)! \\
 &=&\rm f(n\!+\!1) \end{eqnarray}$$
That completes the proof using the theorem. This method works quite widely for inductive proofs involving sums and products. You can find many more examples of telescopy and related results in other answers here.