Hint for prove $\forall n \in \Bbb N \sum _{i=1}^{n} \frac {1}{i!} \le 2 - \frac{1}{2^{n-1}}$? I'm trying to prove the following inequality $\forall n \in \Bbb N$:
$$\sum _{i=1}^{n} \frac {1}{i!} \le 2 - \frac{1}{2^{n-1}}$$
I'm doing it by induction. It's true for $P(1)$.
So now I want to prove it for $P(n+1)$.
$$\sum _{i=1}^{n+1} \frac {1}{i!} \le 2 - \frac{1}{2^{n}}$$
I assume the inductive hypothesis for $P(n)$: $\sum _{i=1}^{n} \frac {1}{i!} \le 2 - \frac{1}{2^{n-1}}$ I'm left with:
$$\sum _{i=1}^{n} \frac {1}{i!} + \frac{1}{(n+1)!} \le 2 - \frac{1}{2^{n}}$$
I'm not assure what step to take next, I should bound it somehow but it's not clear to me how. If anyone could suggest me the next move... thanks.

Hint:
It will be sufficient to show that $$\frac{1}{(n+1)!}\le\frac{1}{2^{n-1}}-\frac{1}{2^n}=\frac{1}{2^n}$$ which follows easily from the fact
$$2^n\le (n+1)!\qquad n\ge 1\text{ integer}\tag{1}$$
Inequality $(1)$ holds since $2\le 1\cdot 2$ and $2< i+1$ for all $1<i\le n$.
