Proving $\sum_{n=0}^\infty\frac{x^n}{n!}>0$ for all $x\in\Bbb R$ without invoking any Taylor series knowledge base. While studying basic Taylor expansions, I came across this expansion of $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}.$$We already know this function is positive for all $x$. But let us assume we don't know about the Taylor expansion , and try to prove this in the expanded form only, how to will we prove it ?
It's already clearly  visible for $ x > 0$ as well as for $ -1<x<0 $ . But how to prove it for $x<-1$ ? I have tried much but not able to build up a strong convincing proof.
Any help would be highly appreciated. thanks!
P.S : prove it 'WITHOUT' using $e^{x}$ function.

To elaborate on some of the hints in the comments, let $x,y \in \mathbb{R}$. Then:
\begin{align*}
e^{x+y} &= \sum_{n=0}^{\infty}\frac{(x+y)^n}{n!} \\
&= \sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{n}{k}\frac{x^ky^{n-k}}{n!} \\
&= \sum_{n=0}^{\infty} \sum_{k=0}^{n}\frac{x^k}{k!}\frac{y^{n-k}}{(n-k)!} \\
&= 1 + \left(x+y\right) + \left(\frac{x^2}{2!}+xy + \frac{y^2}{2!}\right) + \left(\frac{x^3}{3!} + \frac{x^2y}{2!} + \frac{xy^2}{2!} + \frac{y^3}{3!}\right) + \ldots \\
&= \left(1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\right) + y\left(1+x+\frac{x^2}{2!}  + \ldots\right) + \frac{y^2}{2!}\left(1+x + \ldots\right) + \ldots \\
&= \left(1+y+\frac{y^2}{2!} + \frac{y^3}{3!} + \ldots\right)\left(1+x+\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\right) \\
&= e^{x}e^{y}
\end{align*}
This factoring trick is what people in the comments are calling the Cauchy Product. I thought that seeing the algebra done by hand (sort of) might convince you it's not a crazy sophisticated trick. It follows easily from this identity that $e^{-x} = 1/e^{x} > 0$.
