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.
 A: I'll use the Kronecker delta, so the $0^0=1$ we use for an $x^0$ term in Taylor series gives $0^n=\delta_{n0}$ for integers $n\ge0$. Let $[x^n]f(x)$ denote the $x^n$ coefficient in $f(x)$. As @leoli hints,$$[x^n]\sum_{k,\,l\ge0}\frac{x^k(-x)^l}{k!l!}=\sum_{k=0}^n\frac{1}{k!(n-k)!}=\frac{1}{n!}\sum_{k=0}^n\binom{n}{k}(-1)^k=\frac{0^n}{n!}=\delta_{n0},$$so $\sum_{k,\,l\ge0}\frac{x^k(-x)^l}{k!l!}=\sum_n\delta_{n0}x^n=1$ and, for $x<0$,$$\sum_l\frac{(-x)^l}{l!}>0\implies\sum_k\frac{x^k}{k!}=\frac{1}{\sum_l\frac{(-x)^l}{l!}}>0.$$
A: 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 $e^{-x} = 1/e^{x} > 0$.
A: Here is a proof that uses only that $f'=f$, a fact that is easy to prove from the Taylor expansion of $f$.
Letting
$$
  f(x)=\sum_{n=0}^\infty\frac{x^n}{n!},
  $$
suppose by contradiction that there exists $x\in \mathbb R$, such that $f(x)<0$, and let
$$
  a= \sup\{x\in \mathbb R: f(x)<0\}.
  $$
Since $f$ is positive on $\mathbb R_+$, it is clear that $a$ is well defined.  Moreover, since $f$
is continuous, we have that $f(a)=0$.
Observing that $f$ is nonzero on the interval $(a,\infty )$, we see that
$$
  g(x) = \frac{f(x+1)}{f(x)},
  $$
is well defined on that interval.  Moreover,  using that $f'=f$,  one has for every $x$ in $(a,\infty )$ that
$$
  g'(x) = \frac{f'(x+1)f(x) - f(x+1)f'(x)}{f(x)^2} =
  \frac{f(x+1)f(x) - f(x+1)f(x)}{f(x)^2} = 0,
  $$
so $g\equiv c$, for some constant $c$.
For every $x$ in $(a,\infty )$, one then has that
$f(x+1) = c f(x)$, whence
$$
  f(a+1) = \lim_{x\to a^+}f(x+1) = \lim_{x\to a^+}cf(x) = cf(a) =0,
  $$
contradicting the choice of $a$.
