Bernoulli's Inequality for $-1 \leq x\leq 0$

As regards your original goal, there is a shorter way. We have that $$e^x-1-x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n-1-x=\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}.$$ Hence, for $x\in [-1,1]$, $$|e^x-1-x|=\left|\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}\right|\leq x^2\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{|x|^{k-2}}{n^k}\\\leq x^2\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{1}{n^k} \leq x^2\sum_{k=2}^{\infty}\frac{1}{k!}\leq e x^2$$ and the given limit follows as $x\to 0$ by the squeeze theorem.

Along the same argument we show that for $x\in [-1,1]$, $$\left|e^x-\sum_{k=0}^{n}\frac{x^{k}}{k!}\right|<e|x|^{n+1}$$ which implies that $e^x=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$.

Bernoulli inequality for $-1<x<0$: $$e^x-1-x=\sum_{k=2}^{\infty}\frac{x^{k}}{k!}=\sum_{k=1}^{\infty}\frac{x^{2k}}{(2k)!}\underbrace{\left(1+\frac{x}{2k+1}\right)}_{\geq 0}\geq 0.$$