Proving a lemma for the irrationality of $e$ Define
$$I_n = \int_0^1 e^tt^ndt$$ where $n$ is a non-negative integer.
In a related question here, I asked how $e$ can be proven based on the following definitions and results:
$$I_{n+1} = e - (n + 1)I_n$$
$$I_n = (-1)^{n + 1}n! + e\sum_{r = 0}^n (-1)^r\frac{n!}{(n-r)!}$$
$$\frac{1}{n+1}\le I_n < \frac{e}{n} ~~~\text{for all $n \ge 1$}$$
The thing is, how would one go about proving the third inequality? I know that $e^{-1} = \sum_{k=0}^{n} \frac{(-1)^{k}}{k!}$ but I t doesn't seem to be of much use.

Notice that if $n \geq 1$, then $e^tt^n \leq e^1t^n$ for all $t \in [0, 1]$. This implies that:
$$\int_0^1 e^tt^ndt \leq e\int_0^1 t^ndt = \frac{e}{n+1} < \frac{e}{n}$$
Hence, $\displaystyle I_n < \frac{e}{n}$.
