Prove that $\sum_{i=1}^{n} i \times i! = (n+1)! - 1$ by induction \begin{align*}
\sum_{i = 1}^{k + 1} i(i!) & = \sum_{i = 1}^{k} i(i!) + (k + 1)(k + 1)!\\
                           & = (k + 1)! - 1 + (k + 1)(k + 1)! & \text{by the induction hypothesis}\\
                           & = (1 + k + 1)(k + 1)! - 1\\
                           & = (k + 2)(k + 1)! - 1\\
                           & = (k + 2)! - 1
\end{align*}
I have a question from this post solving the problem
Prove by induction that $\sum_{i=1}^n i!\times i=(n+1)!-1$ for all $n\in \mathbb{N}$
How does the person go from 
$                           = (k + 2)(k + 1)! - 1\\
                           = (k + 2)! - 1$
at the very end? I don't understand how the permutation of $(k+1)!$ and (k+2) are able to combine into $(k+2)!$

It is because, by definition, $n! = n(n-1)!$ and $0! = 1$. Just take $n=k+2$.
