Prove that $\alpha_{m}=\sum_{n=1}^{\infty}\frac{1}{(n!)^{m}}$ is irrational for all $m\geq 1$.

(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, $\alpha_m \not \in \mathbb Q.$