Why does $ \lim_{n \to \infty} (\frac{1-e^{\frac{-z(e-1)}{n}-1}}{1-e^{-1}})^n = e^z$?

Question

As part of a probability calculation (convergence of a sequence of CDF's)I need to calculate $$ \lim_{n \to \infty} (\frac{1-e^{\frac{-z(e-1)}{n}-1}}{1-e^{-1}})^n$$ Wolframalpha says that $$ \lim_{n \to \infty} (\frac{1-e^{\frac{-z(e-1)}{n}-1}}{1-e^{-1}})^n = e^z.$$ I can't figure out why... can you?

Answer 1

$$ \lim_{n \to \infty} \left(\frac{1-e^{\frac{-z(e-1)}{n}-1}}{1-e^{-1}}\right)^n=\lim_{n \to \infty} \left(1+\frac{1-e^{\frac{-z(e-1)}{n}-1}}{1-e^{-1}}-1\right)^n=$$ $$=\lim_{n \to \infty} \left(1+\frac{1-e^{\frac{-z(e-1)}{n}}}{e-1}\right)^{\frac{e-1}{1-e^{\frac{-z(e-1)}{n}}}\cdot\frac{n\left(1-e^{\frac{-z(e-1)}{n}}\right)}{e-1}}=e^{\lim\limits_{x\rightarrow0}\frac{1-e^{-z(e-1)x}}{(e-1)x}}=$$ $$=e^{\lim\limits_{x\rightarrow0}\frac{e^{-z(e-1)x}-1}{-z(e-1)x}\cdot z}=e^z$$