Using the Limit definition to find the derivative of $e^x$
Sometimes one defines $e$ as the (unique) number for which $$\tag 1 \lim_{h\to 0}\frac{e^h-1}{h}=1$$
In fact, there are two possible directions.
$(i)$ Start with the logarithm. You'll find out it is continuous monotone increasing on $\Bbb R_{>0}$, and it's range is $\Bbb R$. It follows $\log x=1$ for some $x$. We define this (unique) $x$ to be $e$. Some elementary properties will pop up, and one will be $$\tag 2 \lim\limits_{x\to 0}\frac{\log(1+x)}{x}=1$$
Upon defining $\exp x$ as the inverse of the logarithm, and after some rules, we will get to defining exponentiation of $a>0\in \Bbb R$ as $$a^x:=\exp(x\log a)$$
In said case, $e^x=\exp(x)$, as we expected. $(1)$ will then be an immediate consequence of $(2)$.
$(ii)$ We might define $$e=\sum_{k=0}^\infty \frac 1 {k!}$$ (or the equivalent Bernoulli limit). Then, we may define $$\exp x=\sum_{k=0}^\infty \frac{x^k}{k!}$$ Note $$\tag 3 \exp 1=e$$
We define the $\log$ as the inverse of the exponential function. We may derive certain properties of $\exp x$. The most important ones would be $$\exp(x+y)=\exp x\exp y$$ $$\exp'=\exp$$ $$\exp 0 =1$$
In particular, we have that $\log e=1$ by. We might then define general exponentiation yet again by $$a^x:=\exp(x\log a)$$
Note then that again $e^x=\exp x$. We can prove $(1)$ easily recurring to the series expansion we used.
ADD As for the definition of the logarithm, there are a few ones. One is $$\log x=\int_1^x \frac{dt}{t}$$
Having defined exponentiation of real numbers using rationals by $$a^x=\sup\{a^r:r\in\Bbb Q\wedge r<x\}$$
we might also define $$\log x=\lim_{k\to 0}\frac{x^k-1}{k}$$
In any case, you should be able to prove that
$$\tag 1 \log xy = \log x +\log y $$ $$\tag 2 \log x^a = a\log x $$ $$\tag 3 1-\dfrac 1 x\leq\log x \leq x-1 $$ $$\tag 4\lim\limits_{x\to 0}\dfrac{\log(1+x)}{x}=1 $$ $$\tag 5\dfrac{d}{dx}\log x = \dfrac 1 x$$
What you want is a direct consequence of either $(4)$ or $(5)$, or of the first sentence in my post.
ADD We can prove that for $x \geq 0$ $$\lim\left(1+\frac xn\right)^n=\exp x$$ from definition $(ii)$.
First, note that $${n\choose k}\frac 1{n^k}=\frac{1}{{k!}}\frac{{n\left( {n - 1} \right) \cdots \left( {n - k + 1} \right)}}{{{n^k}}} = \frac{1}{{k!}}\left( {1 - \frac{1}{n}} \right)\left( {1 - \frac{2}{n}} \right) \cdots \left( {1 - \frac{{k - 1}}{n}} \right)$$
Since all the factors to the rightmost are $\leq 1$, we can claim $${n\choose k}\frac{1}{{{n^k}}} \leqslant \frac{1}{{k!}}$$
It follows that $${\left( {1 + \frac{x}{n}} \right)^n}=\sum\limits_{k = 0}^n {{n\choose k}\frac{{{x^k}}}{{{n^k}}}} \leqslant \sum\limits_{k = 0}^n {\frac{{{x^k}}}{{k!}}} $$
It follows that if the limit on the left exists, $$\lim {\left( {1 + \frac{x}{n}} \right)^n} \leqslant \lim \sum\limits_{k = 0}^n {\frac{{{x^k}}}{{k!}}} = \exp x$$
Note that the sums in $$\sum\limits_{k = 0}^n {{n\choose k}\frac{{{x^k}}}{{{n^k}}}} $$
are always increasing, which means that for $m\leq n$
$$\sum\limits_{k = 0}^m {{n\choose k}\frac{{{x^k}}}{{{n^k}}}}\leq \sum\limits_{k = 0}^n {{n\choose k}\frac{{{x^k}}}{{{n^k}}}}$$
By letting $n\to\infty$, since $m$ is fixed on the left side, and $$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{k!}}\left( {1 - \frac{1}{n}} \right)\left( {1 - \frac{2}{n}} \right) \cdots \left( {1 - \frac{{k - 1}}{n}} \right) = \frac{1}{{k!}}$$
we see that if the limit exists, then for each $m$, we have $$\sum\limits_{k = 0}^m {\frac{{{x^k}}}{{k!}}} \leqslant \lim {\left( {1 + \frac{x}{n}} \right)^n}$$
But then, taking $m\to\infty$ $$\exp x = \mathop {\lim }\limits_{m \to \infty } \sum\limits_{k = 0}^m {\frac{{{x^k}}}{{k!}}} \leqslant \lim {\left( {1 + \frac{x}{n}} \right)^n}$$
It follows that if the limit exists $$\eqalign{ & \exp x \leqslant \lim_{n\to\infty} {\left( {1 + \frac{x}{n}} \right)^n} \cr & \exp x \geqslant \lim_{n\to\infty} {\left( {1 + \frac{x}{n}} \right)^n} \cr}$$ which means $$\exp x = \lim_{n\to\infty} {\left( {1 + \frac{x}{n}} \right)^n}$$ Can you show the limit exists?
The case $x<0$ follows now from $$\displaylines{ {\left( {1 - \frac{x}{n}} \right)^{ - n}} = {\left( {\frac{n}{{n - x}}} \right)^n} \cr = {\left( {\frac{{n - x + x}}{{n - x}}} \right)^n} \cr = {\left( {1 + \frac{x}{{n - x}}} \right)^n} \cr} $$
using the squeeze theorem with $\lfloor n-x\rfloor$, $\lceil n-x\rceil$, and the fact $x\to x^{-1}$ is continuous. We care only for terms $n>\lfloor x\rfloor$ to make the above meaningful.
NOTE If you're acquainted with $\limsup$ and $\liminf$; the above can be put differently as $$\eqalign{ & \exp x \leqslant \lim \inf {\left( {1 + \frac{x}{n}} \right)^n} \cr & \exp x \geqslant \lim \sup {\left( {1 + \frac{x}{n}} \right)^n} \cr} $$ which means $$\lim \inf {\left( {1 + \frac{x}{n}} \right)^n} = \lim \sup {\left( {1 + \frac{x}{n}} \right)^n}$$ and proves the limit exists and is equal to $\exp x$.
First prove that $\lim_{h\to 0}\frac{\ln(h+1)}{h}=1$. The switch of $\ln$, $\lim$ is possible because $f(x)=\ln x$, $x>0$ is continuous.
$$\lim_{h\to 0}\ln\left( (1+h)^{\frac{1}{h}} \right)=\ln\lim_{h\to 0}\left( (1+h)^{\frac{1}{h}} \right)=\ln e = 1$$
Now let $u=e^h-1$. We know $h\to 0\iff u\to 0$.
$$\lim_{h\to 0}\frac{e^h-1}{h}=\lim_{u\to 0}\frac{u}{\ln(u+1)}=1$$
If you have have a proof for $\displaystyle\frac{d\ln x}{dx}$, you can always prove $\displaystyle\frac{de^x}{dx}$ by taking $\displaystyle\frac{d\ln e^x}{dx}$ using chain rule.
First, we know: $$\frac{d\ln e^x}{dx} = \frac{dx}{dx} = 1$$
Second, using chain rule:
$$\frac{d\ln e^x}{dx} = \frac{1}{e^x}\frac{de^x}{dx} $$ Using the fact that we know both sides equal $1$, then:
$$\frac{1}{e^x}\frac{de^x}{dx} = 1 $$ Multiplying both sides by $e^x$ gives:
$$\frac{de^x}{dx}= e^x$$