Prove using induction that $1!\cdot 1+ 2!\cdot2+ 3!\cdot3+ ... +n!\cdot n = (n+1)!-1$

Question

Assume $n = k$:

$$(k+1)! -1 = 1!\cdot 1!\cdot1+...+k!\cdot k$$

To test for $k + 1$, I add $k+1$ to the sum as new element and equate its sum with that of the first $k$ elements with the formula adjusted for $k+1$:

$$(k+1)!-1+(k+1)!(k+1) = ((k+1)+1)! -1$$ $$(k+1)!+(k+1)!(k+1) = (k+1)(k+1)!$$ $$(k+1)! = 0$$

Equation doesn't hold water. What I did do wrong?

Answer 1

First addressing your question on equation doesn't hold water.

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

I'm not sure how you arrived at the right hand side of the equation. You should have $(k+2)! = (k+2)(k+1)!$ if you wish to expand it out.

Now to address the way you are doing your induction. Since you are trying to prove that an equation is true, you should start from the left hand side (LHS) and derive the right hand side (RHS) or at least something to that effect.

The proof should look like:

Let $p(n)$ be the statement $1!\cdot 1+ 2!\cdot2+ 3!\cdot3+ … +n!\cdot n = (n+1)!-1$.

$p(0)$ is true because $1!\cdot 1 = (1+1)!-1$.

Assume $p(n)$ is true, then \begin{align*} \text{LHS} &= 1!\cdot 1+ 2!\cdot2+ 3!\cdot3+ … +n!\cdot n + (n+1)!\cdot (n+1)\\ &= (n+1)!-1 + (n+1)!\cdot (n+1)\\ &= (n+2)(n+1)!-1\\ &= (n+2)! - 1\\ &=\text{RHS} \end{align*}