Here's an outline of a proof : We establish that for any natural integer $n, e = \sum_{k=0}^n \frac 1{k!} + \int_0^1 \frac {(1-t)^n}{n!}e^t dt,$ then that for any positive integer $n, 0 < e - \sum_{k=0}^n \frac 1{k!} < \frac 3{(n+1)!}.$ We then proceed by contradiction. Historical remark : Hermite has proved in 1873 that $e$ is transcendental. That's historically the first "real" number for which we managed to prove the transcendental character.

Maybe can you prove what you claim?

Sure. We will show that first relation by induction. The base case is easy and left to the reader. Let $n \geq 0.$ Let us assume that $e = \sum_{k=0}^n \frac 1{k!} + \int_0^1 \frac {(1-t)^n}{n!} e^t dt.$ An integration by parts yields : $\int_0^1 \frac {(1-t)^n}{n!} e^t dt = \frac 1{(n+1)!} + \int_0^1 \frac {(1-t)^{n+1}}{(n+1)!} e^t dt,$ and so $e = \sum_{k=0}^n \frac 1{(n+1)!} + \int_0^1 \frac {(1-t)^{n+1}}{(n+1)!} e^t dt = \sum_{k=0}^{n+1} \frac 1{k!} + \int_0^1 \frac {(1-t)^{n+1}}{(n+1)!} e^t dt.$ Thus our result is proved by induction. Let $n$ be a positive integer. From the above, $0 < e - \sum_{k=0}^n \frac 1{k!} = \int_0^1 \frac {(1-t)^n}{n!}e^t dt < e \int_0^1 \frac {(1-t)^n}{n!}dt = \frac e{(n+1)!} < \frac 3{(n+1)!}.$ Let us then assume for sake of contradiction that $e$ is rational. Then, there exist $(a, b) \in \mathbb{N}^2 / e = \frac ab.$ Let $n$ be an arbitrary positive integer. From the above, we have $0 < \frac ab - \sum_{k=0}^n \frac 1{k!} < \frac 3{(n+1)!},$ which rewrites $0 < an! - b\sum_{k=0}^n \frac {n!}{k!} < \frac {3b}{n+1}.$ Especially, for $n = 3b,$ we have $0 < a(3b)! - b\sum_{k=0}^{3b} \frac {(3b)!}{k!} < \frac {3b}{3b+1} < 1.$ But this is impossible since $an! - b \sum_{k=0}^{3b} \frac {(3b)!}{k!}$ is an integer. Hence, $e$ is irrational.

However, if we need just the irrationality there is a completely elementary proof. Assume $ e = \frac{p}{q}$ for some $ p, q \in \mathbb{N}$ with $ (p, q) = 1$ and $ q \ge 2$ - it is trivial to check, by an easy estimation, that $ e \in (2, 3)$. This leads to $ p \cdot (q - 1)! = \sum_{k=1}^q \prod_{j = k + 1}^q j + \sum_{k \ge q + 1} \frac{q!}{k!}$, thus $ \sum_{k \ge q + 1} \frac{q!}{k!} > 0$ is an integer. Hence let's keep in mind the relation $ \sum_{k \ge q + 1} \frac{q!}{k!} \ge 1$. On the other hand, $ \sum_{k \ge q + 1} \frac{q!}{k!} = \sum_{t \ge 1} \frac{1}{(q + 1) \cdots (q + t)} \le \frac{1}{q + 1} + \sum_{t \ge 2} \frac{1}{(q + t - 1)(q + t)}$, and by reducing terms, $ \frac{1}{q + 1} + \sum_{t \ge 2} \left( \frac{1}{q + t - 1} - \frac{1}{q + t} \right) = \frac{2}{q + 1} \le \frac{2}{3} < 1$, a contradiction.