An application of Bernoulli's Inequality - relevant to an expression for $e^{x}$ I saw on another post the following proposition.
\begin{equation*}
1 + x + \frac{x^{2}}{2!} + ... + \frac{x^{n}}{n!}
≤ \left( 1 − \frac{x}{n} \right)^{−n}  
\end{equation*}
for every real number $0 < x < n$.  The member posting this proposition said that it can be deduced from Bernoulli's Inequality.  How?  Is the inequality valid if a negative value is substituted for $x$? 

Using the inequality $\ln(1+y) \leqslant y$, we have for $0 \leqslant y < 1$,
$$1+y \leqslant e^y = \sum_{k=0}^{\infty} \frac{y^k}{k!} <  \sum_{k=0}^{\infty} y^k = \frac1{1-y},$$
Take $y = x/n$ with $x > 0$. It follows that for $n$ sufficiently large
$$1 + \frac{x}{n} \leqslant e^{x/n} < \left(1 - \frac{x}{n}\right)^{-1},$$
and
$$\left(1 + \frac{x}{n}\right)^n \leqslant e^x < \left(1 - \frac{x}{n}\right)^{-n}.$$
Hence, for $0 < x < n$
$$1 + x + \frac{x^{2}}{2!} + ... + \frac{x^{n}}{n!}
\leqslant e^x \leqslant \left( 1 − \frac{x}{n} \right)^{−n}.$$
