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.

As Angel's answer notes, an ancillary inequality will be of use here, namely 
$$
2^n\leq(n+1)!,\qquad n\geq1\tag{1}
$$
which may be easily proved as well using induction (here is the meat of the inductive proof for $(1)$):
$$
2^{k+1}=2^k\cdot2\leq(k+1)!\cdot2\leq(k+1)!\cdot(k+2)=(k+2)!.
$$
Hence, 
$$
2^n\leq(n+1)!\Longleftrightarrow\frac{1}{2^n}\geq\frac{1}{(n+1)!}.\tag{2}
$$
Thus, the inductive step of your argument looks like so:
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{i!}&= \sum_{i=1}^k\frac{1}{i!}+\frac{1}{(k+1)!}\tag{by definition}\\[1em]
&\leq 2-\frac{1}{2^{k-1}}+\frac{1}{(k+1)!}\tag{by inductive hypothesis}\\[1em]
&\leq 2-\frac{1}{2^{k-1}}+\frac{1}{2^k}\tag{by $(2)$}\\[1em]
&= 2-\frac{2}{2^k}+\frac{1}{2^k}\tag{rewrite}\\[1em]
&= 2-\frac{1}{2^k}.\tag{simplify}
\end{align}
