Could you please verify whether my attempt is fine or contains logical gaps/errors? Thank you so much!
Lemma: $0<e-\sum_{k=0}^{n} \frac{1}{k !}<\frac{1}{n n !}$ for all $n \in \mathbb N ^+$.
Assume the contrary that $e$ is rational, i.e. $e = p/q$ for some $p,q \in \mathbb Z^+$. Let $x=q!\left(e -\sum_{k=0}^{q} \frac{1}{k !}\right)$. It's easy to verify that $x \in \mathbb Z^+$. On the other hand, $x < q!/(qq!) = 1/q <1$ by Lemma, which contradicts the fact that $1 < e <3$. As such, $e$ is irrational.
Update: As suggested by @lulu, I added a proof of the lemma.
We have $$\begin{aligned}e-\sum_{k=0}^{n} \frac{1}{k !} &= \sum_{k=0}^{\infty} \frac{1}{k !} -\sum_{k=0}^{n} \frac{1}{k !} &&=\sum_{k=n+1}^{\infty} \frac{1}{k !} \\&= \sum_{k=n+1}^\infty \frac{1}{n! \prod_{j=n+1}^k j} &&= \frac{1}{n!} \sum_{k=n+1}^\infty \frac{1}{\prod_{j=n+1}^k j} \\& < \frac{1}{n!} \sum_{k=n+1}^\infty \frac{1}{(n+1)^{k-(n+1)+1}} &&= \frac{1}{n!} \sum_{k=n+1}^\infty \left (\frac{1}{n+1} \right)^{k-n} \\&= \frac{1}{n!} \frac{\frac{1}{n+1}}{1-\frac{1}{n+1}} && = \frac{1}{nn!} \end{aligned}$$
Best Answer
I'm gonna give you one proof that is very similar to what you are looking for.
Let's give a general bound on the remainder :
$e^{x} - \sum\limits_{k=0}^{n} \frac{x^{k}}{k!} = \sum\limits_{k \geq n+1} \frac{x^{k}}{k!}$, so, getting the absolute value $\lvert \sum\limits_{k \geq n+1} \frac{x^{k}}{k!}\rvert \leq \sum\limits_{k \geq n+1} \frac{\lvert x\rvert^{k}}{k!}$
Multiply and divide by $\frac{\lvert x \rvert^{n+1}}{(n+1)!}$
$$\lvert \sum\limits_{k \geq n+1} \frac{x^{k}}{k!} \rvert = \frac{\lvert x \rvert^{n+1}}{(n+1)!} \sum\limits_{k \geq n+1} \frac{(n+1)!}{k!}\lvert x \rvert^{k-n-1} \underset{j=k-n-1}{=} \frac{\lvert x \rvert^{n+1}}{(n+1)!}\sum\limits_{j \geq 0} \frac{(n+1)!}{(n+1+j)!}\lvert x \rvert^{j}$$
Observation :
$(n+1+j)! \geq (n+1)!j! \iff \frac{(n+1+j)!}{(n+1)!j!} = \binom{n+1+j}{j} \geq 1$
So $\frac{j!(n+1)!}{(n+1+j)!} \leq 1$, because $\frac{(n+1)!j}{(n+1+j)!} = \frac{1}{\binom{n+1+j}{j}} \leq 1$
Now :
$$\lvert e^{x} - \sum\limits_{k=0}^{n} \frac{x^{k}}{k!} \rvert \leq \frac{\lvert x \rvert^{n+1}}{(n+1)!}\sum\limits_{j \geq 0} \frac{\lvert x \rvert^{j}}{j!} = \frac{\lvert x \rvert^{n+1}}{(n+1)!} e^{\lvert x \rvert}$$
Now we are able to say that $e$ is irrational.
Let's notice that rationals have the following property,for sufficiently big $n \in \mathbb{N}$,when multiplied by $n!$ they become integers,
Let's prove that $e$ doesn't respect this property :
$$n!\cdot e = [n! \sum\limits_{k=0}^{n}\frac{1}{k!}] + n!\sum\limits_{k\geq n+1} \frac{1}{k!}$$
The first term is an integer because if $k \leq n \rightarrow k! \mid n!$
For the second term let's notice given the bound just prooved that $$n! \cdot \sum\limits_{k \geq n+1} \frac{1}{k!} < n! \cdot \frac{e}{(n+1)!} = \frac{e}{(n+1)}$$
Which is not integers for $n \geq 2$ (Note that definitely belongs to $(0,1)$).
$n=1$ we can easily see that is $1 \cdot e$ would be integer it would follow that multiplied by any number would be integer, but multiplied by any $n \geq 2$ it fails due to what we've just saw.
Tell me if there's anything wrong in the proof or if there's something that doesn't feel right to you.