Prove by Induction: Summation of Factorial (n! * n)

Question

Prove by induction (weak or strong) that:

$$(1! \cdot 1) + (2! \cdot 2) + \cdots + (n! \cdot n) =\sum_{k=1}^nk!\cdot k= (n + 1)! - 1$$

My base case is:

$n = 1$, which is true.

And my Inductive Hypothesis is:

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

After that, I'm trying to show the $(k + 1)$-stage where:

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

Which simplifies to:

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

I see that I can substitute in my Inductive Hypothesis but where I'm stuck is manipulating the LHS to be equal to the RHS after that:

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

Answer 1

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

Answer 2

\begin{align*} (1!\cdot 1) + (2!\cdot 2) + \cdots + (k!\cdot k) + ((k+1)!\cdot(k+1)) &= (k+1)!-1 + ((k+1)!\cdot(k+1)) \\ &= ((k+1)!\cdot(k+2))-1 \\ &= (k+1)!-1. \end{align*}

Answer 3

From here

$$\sum_{k=1}^nk!\cdot k= (n + 1)! - 1$$

we have

$$\sum_{k=1}^{n+1}k!\cdot k=\sum_{k=1}^nk!\cdot k+ (n + 1)!(n + 1) =\\=(n + 1)! - 1+ (n + 1)!(n + 1)=(n+1)!(n+2)-1=(n+2)!-1$$

Answer 4

Alternatively: $$\sum_{k=1}^n k!\cdot k=\sum_{k=1}^n k!\cdot (k+1-1)=\sum_{k=1}^n ((k+1)!-k!)=\\ (2!-1!)+(3!-2!)+\cdots+(n!-(n-1)!)+((n+1)!-n!)=(n+1)!-1!=(n+1)!-1.$$