calculus - limit to prove exponential definition for all values of z

In order to prove that the relation $\lim_{n\rightarrow\infty}(1+\frac{z}{n})^n=\exp(z)$ holds for all values of z, I first prove it for positive z by showing the lower and upper bound as taylor series for $exp(z)$. To extend the proof of this relation to hold for negative z, I need to prove $\lim_{n\rightarrow\infty} (1-\frac{z^2}{n^2})^n =1$, I use the binomial expansion of the series and write it down $\lim_{n\rightarrow\infty} \sum_{k=0}^n \frac{z^{2k}}{n^{2k}}\frac{{n}{!}}{{n-k}{!}{k}{!}}$ and further expand the combinatorial term. How do I prove that the limit of this is 1 ?

Answer

Here is an approach.

$$ e^{{n}\ln(1-z^2/n^2)} = e^{{n}(-z^2/n^2+O(z^4/n^4))} \longrightarrow_{n\to \infty} 1 . $$

We used the Taylor series

$$\ln(1-t) = -t+O(t^2) .$$