<s>[INST] Help with proof by induction, please! I may not be understanding how to work with factorials. 
Prove that for all integers $n \ge 1, 1\cdot1! + 2\cdot2! + 3\cdot3! + \cdots + n\cdot n! = (n+1)! - 1$.

OK, so I verified base case $n = 1: 2! - 1 = 1$
I assumed that for all $k \ge n, P(k) = (k+1)! - 1$
I am unsure how to prove true for $P(k+1)$, though I understand that the first step is to replace all $k$ values with $k+1$ which yields $$1\cdot1! + 2\cdot2! + 3\cdot3! + \cdots + k!\cdot k! + (k+1\cdot k+1)! = (k+2)! - 1$$
What next?
 [/INST] Hint:  If $f(n) = 1*1! + 2*2! + ..... + n*n!$ then $f(n+1) = 1*1! + 2*2! + ..... + n*n! + (n+1)(n+1)! = f(n) + (n+1)(n+1)!$.
Hint 2:   $m!(m+1) = (m+1)!$ and $m!*m + m! {=\over{\text{factor}}} m!(m+1) = m!(m+1)= m!$.

 So if we assume $f(k) = 1*1! + 2*2! + ..... + k*k!= (k+1)! - 1$.

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Then $f(k+1) = f(k) + (k+1)(k+1)! =$

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 $(k +1)! -1 +  (k+1)(k+1)!=(k+1)! + (k+1)(k+1)! - 1 = (k+1)![1+(k+1)]-1=$ 

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$(k+1)![k+2]-1 = (k+2)!-1$ 

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