$\sum_{i=1}^n i\cdot i! = (n+1)!-1$ By Induction

I am trying to prove the following by Mathematical Induction:

$$\sum_{i=1}^n i\cdot i! = (n+1)!-1\quad\text{for all integers $n\ge 1$}$$

My proof by Induction follows:
First prove $P(1)$ is true,

$$\sum_{i=1}^1 i\cdot i! = (1+1)! – 1$$

Then, for all integers $k >= 1$, if $P(k)$ is true then $P(k+1)$ is also true,

$$ \sum_{i=1}^k i\cdot i! = (k+1)! – 1$$
Then, we must show that $P(k+1)$ is true

$$ \sum_{i=1}^{k+1} i\cdot i! = ((k+1)+1)! – 1\\
\sum_{i=1}^{k+1} i\cdot i! = (k+2)! – 1$$

I am currently struggling with the next step to show that $P(k)$ is equal to $P(k+1)$. How would I finish this prove by induction?

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