Proof that $(n+1)m!>(m+1)!−1$ does not hold for $m>n$

Question

I am currently doing a project that involves some work on Liouville's theorem for transcendental numbers and Liouville's constant. I have found a proof that Liouville's constant is transcendental however it come with this note:

"so that $(n+1)m!>(m+1)!−1$ for all sufficiently large m. But this is false for any value of m greater than n (the reader should give a detailed proof of this statement)" - Taken from What is Mathematics? R. Courant, H. Robbins, I. Stewart.

Since I want my project to be as thorough as possible I would like to include this proof however I don't know how to do it. I know that the starting point is $(m+1)!=(m+1)m!$ but not sure what my next steps should be.

Thanks in advance

Answer 1

The inequality does hold for $m=1,n=0$ so you need to exclude that case. Otherwise $m! \gt 1$ and we can say $(m+1)!-1=(m+1)m!-1=(n+1)m!+(m-n)m!-1 \gt (n+1)m!$ because $m-n \ge 1$

Answer 2

For $m=n+p>n$ we have

$$(n+1)m!=(n+1)(n+p)!<(n+p+1)(n+p)!-p(n+p)!=$$$$=(n+p+1)!-p(n+p)!=(m+1)!-p(n+p)!\le(m+1)!−1$$

Answer 3

Hint: Factorize: $$(n+1)m!>(m+1)!−1 \iff \\ m!(n+1-m-1)>-1 \iff \\ m!(n-m)>-1.$$