Proving $n!=n(n-1)!$

Question

Recently, I saw a question like this. But I knew that It's just a concept that $$n!=n(n-1)(n-2)......(3)(2)(1)$$ But the question had made me curious if there is any proof of the statement.

Show that, $$n!=n(n-1)!$$

So after seeing the question my attempt was : $$6!=6.5.4.3.2.1=6(6-1)(6-2)(6-3)(6-4)(6-5)=6(6-1)!$$ Taking n=6 hence proved. But I believe there is some other way to prove it, what's that?

Answer 1

Let's use the mathematical induction.

Claim) $n!=n(n-1)!$.

if $n=1$:

$1!=1 \cdot (1-1)! = 1 \cdot 0! = 1 \cdot 1 = 1.$

Let's assume that the claim is right until $n=k$.

if $n=k+1$:

$(k+1)! = (k+1)(k)(k-1)(k-2)\cdots (2)(1) = (k+1)\{(k)(k-1)(k-2)\cdots(2)(1)\} = (k+1) \cdot k!$

$\therefore n! = n(n-1)!$.