I am stuck halfway while using mathematical induction to prove that $\left(r+1\right)!-r!=r\left(r!\right)$
I know that the first step is to prove the basis is true, therefore, I substituted r=1 into the equation, proving $1$ = $1$ therefore $LHS$ = $RHS$.
The second step is to assume $\left(r+1\right)!-r!=r\left(r!\right)$ is true for $r=k$ therefore, $\left(k+1\right)!-k!=k\left(k!\right)$ and thus, I need to use this assumption to prove that $k+1$ is true. But I do not know how to prove $k+1$ is true. When I substitued $r=k+1$, I got $\left(k+2\right)!-(k+1)!=(k+1)\left((k+1\right))!$.
How do i proceed from here?
Best Answer
$$\begin{array}{l}\left(r+1\right)!-r!=\left(r+1\right)r!-r!=\left(\left(r+1\right)-1\right)r!=r\left(r!\right)\\\end{array}$$
by substituting $$r=k+1$$
$$ \begin{array}{l}\left(k+1+1\right)!-\left(k+1\right)!\;\\=\;\left(k+2\right)!-\left(k+1\right)!\;\\=\left(k+2\right)\left(k+1\right)!-\left(k+1\right)!\;\\=\left(k+1\right)!\left(k+2-1\right)\\=\left(k+1\right)!\left(k+1\right)\end{array}$$ $$\begin{array}{l}=\left(k+1\right)\left(k+1\right)!\\=r\left(r!\right)\end{array}$$