Given the series definition of $e^x$:
$$e^x=\sum^\infty_{n=0}\frac{x^n}{n!}$$
How does one show that $e^x\ge0$ for $x\in\mathbb{R}$?
One way is to show that
$$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n$$
If $n>|x|$, $1+\frac{x}{n}>0$. By induction one can show $x^n>0$ for $x>0$, so
$$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n>0^n=0$$
However, I was wondering if there was a more direct way to prove this, as proving the equivalence of the series and limit can be quite tedious.
For every $n\in\mathbb{Z}_0$, $x\in\mathbb{R}_+$, $\frac{x^n}{n!}>0$, so $e^x\ge0$ for $x\in\mathbb{R}_+$. Also $e^0=1$ (as the other terms equal $0$).
The tricky part is in showing $e^x\ge0$ for $x<0$. My attempt was this:
Let $f(x)=e^x$ for notational simplicity. Suppose for the sake of contradiction, $f(x_0)<0$ where $x_0<0$. $f'(x)=f(x)$, so $f'(x_0)<0$. As $e^x$ is continuous, there exists a $\delta_0>0$ such that $|x-x_0|\le\delta_0\Rightarrow f(x)\le0$. Let $x_1=x_0+\delta_0$. So $\forall x\in[x_0, x_1]$, $f'(x)<0$. By MVT, $f(x_1)-f(x_0)=\delta_0f'(c)$ for some $c\in[x_0, x_1]$, and as shown earlier $f'(c)\le0$. As $\delta_0>0$,
$$f(x_1)-f(x_0)\le0\iff f(x_1)\le f(x_0)<0$$
We then construct $\delta_1>0$, and $x_2=x_1+\delta_1$ such that $f(x_2)\le f(x_1)<0$, and so on, ad infinitum.
The issue with this proof, however, is that I am unable to say anything about the size of $\delta_0, \delta_1, \delta_2, \cdots$. It is possible that
$$\sum^\infty_{n=0}\delta_n=L$$
for some finite $L$ (e.g. if $\delta_0=1, \delta_1=\frac{1}{2}, \delta_2=\frac{1}{4}$ e.t.c.). Through induction one can show $x_m=x_0+\sum^{m-1}_{i=0}\delta_i$, so
$$\lim_{j\to\infty}x_j=x_0+\sum^\infty_{n=0}\delta_n=x_0+L$$
And it is possible $x_0+L<0$, meaning there is no contradiction.
How can I fix this proof (or if it is unsalvageable, what would be a working proof?)
Hint : You can use the three following (and easy to prove, with the series definition) facts :
$x \mapsto e^x$ is continuous,
$x \mapsto e^x$ does not vanish,
$e^0=1$