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?

  • 3
    $\begingroup$ Why do you need induction? $$ (r+1)!-r! = (r+1)r!-r! = ((r+1)-1)r! = r (r!) $$ $\endgroup$ – MisterRiemann Sep 2 '18 at 8:11
  • $\begingroup$ I was told to use the induction method to prove to ensure that we know the concept of mathematical induction. I showed the above mentioned method but I was told even though I proved it, it was not the method wanted $\endgroup$ – deviljones Sep 2 '18 at 8:14
  • 2
    $\begingroup$ You can do what Sobi did with $(r + 1)! - r!$ for $r = k + 1$ to complete the induction step. $\endgroup$ – N. F. Taussig Sep 2 '18 at 8:16
  • $\begingroup$ Don't understand why I didnt think of it... thanks! $\endgroup$ – deviljones Sep 2 '18 at 8:38


by substituting $$r=k+1$$

$$ \begin{array}{l}\left(k+1+1\right)!-\left(k+1\right)!\;\\=\;\left(k+2\right)!-\left(k+1\right)!\;\\=\left(k+2\right)\left(k+1\right)!-\left(k+1\right)!\;\\=\left(k+1\right)!\left(k+2-1\right)\\=\left(k+1\right)!\left(k+1\right)\end{array}$$ $$\begin{array}{l}=\left(k+1\right)\left(k+1\right)!\\=r\left(r!\right)\end{array}$$


As noticed in the comments we don't need induction since we have


and the given identity holds.

  • $\begingroup$ Yes we don't have to, but I was explicitly told to use the induction method to prove instead. Nevertheless, thanks! $\endgroup$ – deviljones Sep 2 '18 at 8:43
  • $\begingroup$ @deviljones Yes we can of course but the induction step reduces to the same identity which is trivial. $\endgroup$ – gimusi Sep 2 '18 at 8:50

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