Proof, that $n! < m!$ when $n<m$

Question

For another proof I just assumed, that $n! < m! \;$ for all $\;n<m$. And I mean it is kind of obvious that this is the case, but then again I should be able to prove it. However I have no idea where to start there. Can someone give me a hint (or the solution)? Thanks in advance!

Answer 1

If $n<m$, $$n!=1\times2\times...\times(n-1)\times n\qquad\qquad\text{and,}$$ $$m!=\underbrace{1\times2\times...\times (n-1)\times n}_{n!}\times (n+1)\times...\times(m-1)\times m\\=n!\times(n+1)\times...\times(m-1)\times m.$$

Hence,$$\frac{m!}{n!}=(n+1)\times...\times(m-1)\times m>1\Rightarrow m!>n!$$ since, $m>n\ge1\Rightarrow(n+1),...,(m-1)>1$.

Answer 2

Hint: Notice that if $n<m$, then $$m!=m\cdot (m-1)\cdot(m-2)\ldots\cdot n\cdot(n-1)\ldots \cdot1 = m\cdot (m-1)\cdot\ldots\cdot n!$$