Proving $\left(1- \frac{x}{n+1}\right)^n > \frac{1}{\sum_{i=0}^{n} \frac{x^i}{i!}}$ for $x\in(0,1)$ (by induction?) 
I'm trying to prove that for any $x \in (0,1)$, $$\left(1- \frac{x}{n+1}\right)^n > \frac{1}{\sum_{i=0}^{n} \frac{x^i}{i!}} \,\
,\,\,\,\text{for all $n$}$$

Proceeding by induction, the result is true when $n=1$ (by considering the product $(1- \frac{x}{2}) (1+x) >1$ ). Assuming the result is true when $n=k-1.$ When $n=k,$ $$ \left(1- \frac{x}{k+1}\right)^k\, \sum_{i=0}^{k} \frac{x^i}{i!}$$
I'm stuck here as I'm not sure how to use the induction assumption for $k-1$.
-update-
It seems $\left(1-\frac{x}{n+1}\right)^n$ is decreasing in $n$ (because $0<x<1$). Is there a way to use this fact to obtain the desired inequality?
 A: Just for fun, the opposite inequality (almost). By Taylor's formula
$$
\exp(x)= 1+x + \frac x2 \exp(c)
$$
with $c$ between $0$ and $x$. As a consequence,
$$
\exp u >  1+u
$$
for any $u\neq0$. Applying this result to $u=-\frac{x}{n+1}$ gives for any $x\in(0,n)$
$$
\exp\left(-\frac x{n+1}\right) >  1-\frac{x}{n+1} > 0,
$$
and
$$
\exp\left(-x\frac{n}{n+1}\right)>\left(1-\frac{x}{n+1}\right)^{n}.
$$
Now,
$$
\exp\left(-x\frac{n}{n+1}\right) = \frac{1}{\exp\left(x\frac{n}{n+1}\right)}=\frac{1}{\displaystyle \sum_{k=0}^\infty \left(\frac{xn}{n+1}\right)^k \frac{1}{k!}} < \frac{1}{\displaystyle \sum_{i=0}^n \left(\frac{xn}{n+1}\right)^i \frac{1}{i!}},
$$
so
$$
\left(1-\frac{x}{n+1}\right)^{n} < \frac{1}{\displaystyle \sum_{i=0}^n \left(\frac{xn}{n+1}\right)^i \frac{1}{i!}}.
$$
Now the inequality you wanted. When $x\in(0,1)$,
$$
\frac{1}{\displaystyle 1+\frac x{n}} <  1-\frac x{n+1}.
$$
Indeed,
$$
\left(1+\frac xn \right)\left(1-\frac x{n+1} \right) = 1+\frac{x-x^2}{n(n+1)}>1.
$$
So
$$
\left(1-\frac{x}{n+1}\right)^{n} > \frac{1}{\left(1+\frac{x}{n}\right)^{n}} $$
and we only need to show that
$$
\left(1+\frac{x}{n}\right)^{n} \leq \sum_{i=0}^n \frac{x^i}{i!}
$$
Now
$$
\left(1+\frac{x}{n}\right)^{n} =\sum_{i=0}^n \frac{x^{i}}{i!} \frac{n!}{n^i(n-i)!},
$$
and
$$
\frac{n!}{n^i(n-i)!} = \frac{n(n-1)\ldots{(n-i+1)}}{n^i} <1
$$
so that's that.
