Prove that $\lim_\limits{x\to 0}{\frac{e^x-1}{x}}=1$ without derivatives 
Prove that $\lim_\limits{x\to 0}{\frac{e^x-1}{x}}=1$.

I currently know only one approach (using L'Hopital 's Rule and derivatives) as follows:
$$\lim_\limits{x\to 0}{\frac{e^x-1}{x}}=\lim_\limits{x\to 0}{\frac{\left(e^x-1\right)'}{x'}}=\lim_\limits{x\to 0}{\left(e^x\right)}=e^0=1$$
Here I ask for other proofs than those, preferably neither using derivatives in any way nor using Taylor, etc.
For the purposes of this post, I define the exponential by any of the following limits:
$$e^x
=\lim_\limits{n\to +\infty}{\left( 1+\frac{x}{n}\right)^n}=\lim_\limits{n\to +\infty}{\left[ \left( 1+\frac{1}{n}\right)^{n\cdot x}\right] }=\left[ \lim_\limits{n\to +\infty}{\left( 1+\frac{1}{n}\right)^n}\right] ^x.$$
Note: An approach for $\lim_\limits{x\to 0^+}{(x\ln x)}$ without using derivatives can be found here.

We can assume the "Important limit" $\lim_{x\rightarrow 0}(1+x)^{1/x}=e$. Then
\begin{align*}
\lim_{x\rightarrow 0} \frac{e^x-1}{x} &= \lim_{y\rightarrow 0}\frac{y}{\ln (1+y)}\\
&= \lim_{y\rightarrow 0}\frac{1}{\ln \left((1+y)^{1/y}\right)}\\
&= 1.
\end{align*}
