<s>[INST] Prove by Mathematical Induction: $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$ 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$

 [/INST] Your LHS may not look much like your RHS yet, but that's because you haven't finished getting it into the simplest possible form.  You have $(k+1)k! - 1 + (k+1)^2 k!$.  You're looking to get something minus $1$, so that's somewhat promising.  Now what factors do the other two terms (the ones involving $k$) have in common?
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