We have the series$$ e^{-1} = \sum_{k=0}^\infty {(-1)^k \over k!}$$ Note that this is an alternating series and that the magnitudes of the terms decrease. Hence, for every integer $n > 0$ , we have the bound$$ 0 < \left| \sum_{k=0}^{n} {(-1)^k \over k!} - e^{-1} \right| < {1 \over (n+1)!},$$ by the Leibniz' estimate for alternating series. Assume that $e = n/m$ , where $m$ and $n$ are integers and $n > 0$ . Then we would have$$ 0 < \left| \sum_{k=0}^{n} {(-1)^k \over k!} - {m \over n} \right| < {1 \over (n+1)!} .$$ Multiplying both sides by $n!$ , this would imply$$ 0 < \left| \sum_{k=0}^{n} {(-1)^k n! \over k!} - m (n-1)! \right| < {1 \over n+1} ,$$ which is a contradiction because every term in the sum is an integer, but there are no integers between $0$ and $1/(n+1)$ .