Prove that $\alpha_{m}=\sum_{n=1}^{\infty}\frac{1}{(n!)^{m}}$ is irrational for all $m\geq 1$. I came across this question and went through the proof that $e$ is irrational with a few minor tweaks. I hope someone can have a look through it and hopefully check it or tidy it up. Thanks.
Proof: To begin, note that for $m\geq 1$:
$$\alpha_{m} \,=\, 1 + \sum_{n=2}^{\infty}\frac{1}{(n!)^{m}} \,>\, 1 $$
and 
$$\alpha_{m} \,=\, -1 + \sum_{n=0}^{\infty}\frac{1}{(n!)^{m}} \,<\, -1 + \sum_{n=0}^{\infty}\frac{1}{n!} \,=\, -1 + e \,<\, 2 .$$
Thus $1<\alpha_{m}<2$ for all $m\geq 1$ and so $\alpha_{m}\notin\mathbb{Z}$. Now assume (for contradiction) that $\alpha_{m}\in\mathbb{Q}$:
$$\exists\, p,q\in\mathbb{N} \quad\text{with}\quad q\,>\,1\quad : \quad \alpha_{m}=\frac{p}{q}.$$
Since $p,q\in\mathbb{N}$ and $m\geq 1$: 
$$(q!)^{m}\alpha_{m} \;=\; (q!)^{m}\cdot \frac{p}{q} \;=\; q!\,(q!)^{m-1}\cdot \frac{p}{q} \;=\;  (q-1)!(q!)^{m-1}p$$
and hence $(q!)^{m}\alpha_{m}\in\mathbb{Z}$. Now
$$\begin{align}
(q!)^{m}\alpha_{m} &\;=\; (q!)^{m}\sum_{n=1}^{\infty}\frac{1}{(n!)^{m}} \;=\;
 (q!)^{m}\left(\sum_{n=1}^{q}\frac{1}{(n!)^{m}} + \sum_{n=q+1}^{\infty}\frac{1}{(n!)^{m}}  \right)  \\[0.2cm]
&\;=\; \sum_{n=1}^{q}\left(\frac{q!}{(n!)}\right)^{m} + \sum_{n=q+1}^{\infty}\left(\frac{q!}{(n!)}\right)^{m} \;=\; N + \sum_{n=q+1}^{\infty}\left(\frac{q!}{(n!)}\right)^{m}
\end{align}$$
for some $N\in\mathbb{Z}$, since $n!|q!$ for each $n\leq q$. In particular, since the series in the last term is positive, we have the bound:
$$N \;<\; (q!)^{m}\alpha_{m} \;=\; N + \sum_{n=q+1}^{\infty}\left(\frac{q!}{(n!)}\right)^{m}.$$
Considering the sum:
$$\begin{align}
\sum_{n=q+1}^{\infty}\left(\frac{q!}{n!}\right)^{m} &\;=\; \left(\frac{q!}{(q+1)!}\right)^{m} + \left(\frac{q!}{(q+2)!}\right)^{m} + \left(\frac{q!}{(q+3)!}\right)^{m} + \cdots \\[0.2cm]
&\;=\; \frac{1}{(q+1)^{m}} + \frac{1}{(q+1)^{m}(q+2)^{m}} + \frac{1}{(q+1)^{m}(q+2)^{m}(q+3)^{m}}+ \cdots \\[0.2cm]
&\;<\; \frac{1}{(q+1)^{m}} + \frac{1}{(q+1)^{m}(q+1)^{m}} + \frac{1}{(q+1)^{m}(q+1)^{m}(q+1)^{m}} + \cdots \\[0.2cm]
&\;=\; \frac{1}{(q+1)^{m}} + \frac{1}{(q+1)^{2m}} + \frac{1}{(q+1)^{3m}} + \cdots.
\end{align}$$
This is a geometric series with ratio $0<\frac{1}{(q+1)^{m}}<1$ for all $m\geq 1$. Hence
$$\sum_{n=q+1}^{\infty}\left(\frac{q!}{n!}\right)^{m} \;=\; \frac{\frac{1}{(q+1)^{m}}}{1-\frac{1}{(q+1)^{m}}} \;=\; \frac{1}{(q+1)^{m}-1} \;<\; 1$$
for all $m\geq 1$ since $(q+1)^{m}>1$. Thus we have arrived at:
$$N \;<\; (q!)^{m}\alpha_{m} \;=\; N + \sum_{n=q+1}^{\infty}\left(\frac{q!}{n!}\right)^{m} \;<\; N + 1.$$
But no integer exists in the interval $(N,N+1)$ and hence we have a contradiction; our assumption must be false and $\alpha_{m}\notin\mathbb{Q} \; \blacksquare$. 

(I).Lemma. For $x\in \mathbb R,$ if for each $\epsilon >0$ there exist $a,b\in \mathbb Z$ with $0<|x-a/b|<\epsilon /|b|,$ then $x \not \in \mathbb Q.$ Proof: If $x=c/d$ with $c,d\in \mathbb Z$ then $$0<|x-a/b|<\epsilon /|b|\implies 0<|c/d-a/b|<\epsilon /|b|\implies$$  $$\implies 0<|cb-ad|<\epsilon |d|\implies$$ $$\implies 1\leq |cb-ad|<\epsilon |d|\implies$$ $$\implies \epsilon >1/|d|.$$ 
(II). For $k\geq 2$ we have $1+\sum_{n=2}^k(n!)^{-m}=a_k/b_k$ where $a_k\in  \mathbb Z$ and $b_k=(k!)^{-m}.$ We have   $$0<\alpha_m -a_k/b_k $$ and it is easy to see that $$\lim_{k\to \infty} b_k(\alpha_m-a_k/b_k)=\lim_{k\to \infty}\sum_{n=k+1}^{\infty}(k!/n!)^m=0$$  because $0<(k!/n!)^m\leq (k+1)^{-(n-k)}$ when $n>k\geq 2$.
So by the above lemma, $\alpha_m \not \in \mathbb Q.$
