Prove the series $ \sum_{n=1}^\infty \frac{1}{(n!)^2}$ converges to an irrational number

How can one prove that the series $\displaystyle \sum\limits_{n=1}^\infty \frac{1}{(n!)^2}$ converges to an irrational number? There's no need to use Taylor expansion, integrals or any advanced/professional techniques. It can be proved using only basic techniques. My first attempt was to assume that the series does converge to a rational number $ \frac{p}{q}$ and then it follows by definition that for every positive epsilon there exists an integer k such that $ {\sum_{n=1}^{k} \frac{1}{(n!)^2} -\frac{p}{q}}<=\epsilon$ I tried to get a contradiction but I failed.

Solution 1:

Here is a tiny proof that $e$ is irrational. Your question can be solved the same way.

Solution 2:

Hint: if $x=\frac pq$, what can you say about $x(q!)^2$?

Another Hint: what can you say about $(q!)^2\sum\limits_{n=1}^q\frac1{(n!)^2}$? what about $(q!)^2\sum\limits_{n=q+1}^\infty\frac1{(n!)^2}$?