The Wikipedia proof of the irrationality of $e$ is a little more complicated than necessary, because one has to do a bit of work to estimate the "tail." (We assume familiarity with the Wikipedia proof, which is the usual one.) We prove the same result, using the same basic idea, but with a little twist.
From the series for $e^x$, evaluated at $x=-1$, we can see that $e^{-1}=\sum_2^\infty (-1)^n\frac{1}{n!}=\frac{1}{2!}-\frac{1}{3!} +\frac{1}{4!}-\frac{1}{5!}+\cdots.$
Suppose that $e$ is rational. Then $e^{-1}$ is also rational, say $e^{-1}=a/b$, and therefore $b!\, e^{-1}$ is an integer $N$. But $N=b!\,e^{-1}=H+T$, where $H=b!\:\sum_{2}^b (-1)^n \frac{1}{n!}$ and $T=b!\:\sum_{b+1}^\infty (-1)^n\frac{1}{n!}.$ Note that $H$ is an integer. If we can show that the "tail" $T$ is not an integer, we will have a contradiction.
By a standard fact about alternating series, the error when we truncate an alternating series has absolute value less than the absolute value of the first neglected term, and the same sign as the first neglected term.
If we truncate the series for $e^{-1}$ at the term $(-1)^b \frac{1}{b!}$, the error is therefore non-zero and has absolute value less than $\frac{1}{(b+1)!}$. It follows that $0<|T| < \frac{b!}{(b+1)!}<1,$ and therefore $T$ cannot be an integer.
Comment: In the proof on Wikipedia, we bound the tail by comparing with a geometric series. This is certainly not difficult! But in the setting of a standard calculus course, the basic facts about alternating series are always discussed, and one might as well exploit that.