Show that $e=1+1/1!+1/2!+1/3!+…$ is an irrational number.
Hint: show that, for all positive integers $p$, $0<p![e−(1+1/1!+…+1/p!)]<1$. Then conclude that $e$ cannot be a ratio of two integers q/p.
Attempt #1
I am trying to utilize the hint: the base case for a proof by induction on $p$ is easy to show, but I am having some trouble carrying out the induction step.
Attempt # 2
Then I thought that instead of going for induction, perhaps use the Lagrange-remainder instead, for the function $f(x) = e^x$, evaluated at $x=1$.
Then, after expansion of $p$ terms of $e^x$, about $x=0$, and evaluated at $x=1$, I have that the hint gives \begin{align} p!\left[\frac {f^{(p+1)}(\xi)}{(p+1)!} (x-x_0)^{p+1}\right] &=p!\left[\frac {e^{\xi}}{(p+1)!} (1-0)^{p+1}\right]\\ &=p!\left[\frac {e^{\xi}}{(p+1)!}\right]\\ &=\left[\frac {e^{\xi}}{(p+1)}\right] \end{align}
for some $\xi$ $\in$ (0,1). But now I am stuck. Am I on the right track with using the Lagrange remainder, or should I stick with an induction proof, or perhaps try something else with the hint given?
Thanks,