I was looking through in stack math and got this answer: Prove that $e$ is irrational by Yiorgos S. Smyrlis. (This answer is copied below). Since I cannot comment I can only ask here.
Here is the information provided.
Hints.
We first show that $2<\mathrm{e}<3$ (see below), and hence $\mathrm{e}$ is not an integer.
Next, following up OP's thought, assuming $\mathrm{e}=a/b$, we multiply by $b!$ and we obtain $$ \sum_{k=0}^\infty \frac{b!}{k!}=a\cdot (b-1)! \tag{1} $$ The right hand side of $(1)$ is an integer.
The left hand side of $(1)$ is of the form $$ \sum_{k=0}^b \frac{b!}{k!}+\sum_{k=b+1}^\infty \frac{b!}{k!}= p+r. $$ Note that $p=\sum_{k=0}^b \frac{b!}{k!}$ is an integer, while $$ 0<r=\sum_{k=b+1}^\infty \frac{b!}{k!}=\frac{1}{b+1}+\frac{1}{(b+1)(b+2)}+\cdots<\sum_{k=1}^\infty \frac{1}{(b+1)^k}=\frac{1}{b}<1. $$
Note. The fact that $\mathrm{e}\in (2,3)$ can be derived from the inequalities $$ \left(1+\frac{1}{n}\right)^{\!n}<\mathrm{e}<\left(1+\frac{1}{n}\right)^{\!n+1}, $$ for $n=1$ for the left inequality and $n=5$ for the right inequality.
$$\sum_{k=1}^{∞}\frac{1}{\left(b+1\right)^{k}}=\frac{\frac{1}{\left(b+1\right)}}{1-\frac{1}{\left(b+1\right)}}=\frac{1}{b}<1$$
I used a well-known formula for geometric series, also since $b$ is a positive integer so $\frac{1}{b}$ , but if $b=1$ then $a/b=a$ is strickly between $2$ and $3$ which is a contra since $a$ is supposed to be an integer.
here is a proof of mine which is about irrationaly of $1/e$ which is also useful for proving irrationaly of $e$.