I'm following the proof for the transcendence of $e$ from the book "Transcendental Numbers" by M. Ram Murty and Purusottam Rath. I am struggling to understand the final few lines.
As far as I understand, by the triangle inequality we have
$$ |J|\leq \sum_{k=0}^n |a_k| |I(k,f)|. $$ Now $I(0,f)=0$, so the sum starts from $k=1$. So then
$$ \sum_{k=1}^n |a_k| |I(k,f)|\leq nA\max_{1\leq k\leq n}|I(k,f)| \leq nA\max_{1\leq k\leq n}\{ke^kF(k)\} = n^2Ae^n\max_{1\leq k\leq n}F(k) $$ where $A:=\max\limits_{1\leq k\leq n}|a_k|$.
This is where I'm stumped because I don't know how to bound $F(k)$. I've read some other proofs for help, but they all slightly differ at this step.
Also, where does the "elementary" observation that $e^p\geq p^{p-1}/(p-1)!$ come from? And why is this needed? Can't we just say that the factorial function grows faster than the exponential function so that eventually the inequality $(p-1)!\leq (2n)!^p$ would be false?
Any help would be greatly appreciated.
As regards the bound on $F$, we have that $$F(x)=x^{p-1}(x+1)^p\cdots(x+n)^p$$ and therefore, for $k=1,\dots,n$, $$F(k)k=k^{p}(k+1)^p\cdots(k+n)^p\leq ((2n)!)^p.$$ Hence $$\sum_{k=0}^n|a_k|e^kF(k)k\leq Ae^n\sum_{k=1}^nF(k)k\leq Ae^n\sum_{k=1}^n((2n)!)^p=Ane^n((2n)!)^p.$$
Note that $$e^p=\sum_{k=0}^{\infty}\frac{p^k}{k!}>\sum_{k=0}^{p-1}\frac{p^k}{k!}=1+p+\frac{p^2}{2!}+\dots +\frac{p^{p-1}}{(p-1)!}>\frac{p^{p-1}}{(p-1)!}.$$ The contradiction follows from the fact that $$p^{p-1}e^{-p}\leq Ane^n((2n)!)^p$$ does not eventually hold. Indeed, when $p$ is large enough, the LHS becomes greater than the RHS because $$\lim_{p\to +\infty}\frac{p^{p-1}}{Ane^{p+n}((2n)!)^p}=+\infty.$$ You are right, it is not needed and your argument works as well.