Fundamental lemma. For every sequence of complex numbers $$w_nw_n$$ with a limit $$ww$$ it is true that $$\lim_{n \to \infty} \Bigl(1 + \frac{w_n}{n}\Bigr)^n = \sum_{k=0}^\infty \frac{w^k}{k!}.\lim_{n \to \infty} \Bigl(1 + \frac{w_n}{n}\Bigr)^n = \sum_{k=0}^\infty \frac{w^k}{k!}.$$ Proof. For every $$\varepsilon > 0\varepsilon > 0$$ and sufficiently large index $$KK$$ we have the following estimations: $$\sum_{k=K}^\infty \frac{(|w|+1)^k}{k!} < \frac \varepsilon 3 \quad\mbox{and}\quad |w_n| \le |w|+1.\sum_{k=K}^\infty \frac{(|w|+1)^k}{k!} < \frac \varepsilon 3 \quad\mbox{and}\quad |w_n| \le |w|+1.$$Therefore if $$n \ge Kn \ge K$$ then $$\left|\Bigl(1 + \frac{w_n}{n}\Big)^n - \exp w \right| \le \sum_{k=0}^{K-1} \left|{n \choose k}\frac{w_n^k}{n^k} - \frac{w^k}{k!}\right| + \sum_{k=K}^n{n\choose k} \frac{|w_n|^k}{n^k} + \sum_{k=K}^\infty \frac{|w|^k}{k!}.\left|\Bigl(1 + \frac{w_n}{n}\Big)^n - \exp w \right| \le \sum_{k=0}^{K-1} \left|{n \choose k}\frac{w_n^k}{n^k} - \frac{w^k}{k!}\right| + \sum_{k=K}^n{n\choose k} \frac{|w_n|^k}{n^k} + \sum_{k=K}^\infty \frac{|w|^k}{k!}.$$ The third sum is smaller than $$\varepsilon / 3\varepsilon / 3$$ based on our estimations. We can find an upper bound for the middle one using $${n \choose k} \frac 1 {n^k} = \frac{1}{k!} \prod_{i = 1}^{k-1} \Bigl(1 - \frac i n \Bigr) \le \frac 1 {k!}.{n \choose k} \frac 1 {n^k} = \frac{1}{k!} \prod_{i = 1}^{k-1} \Bigl(1 - \frac i n \Bigr) \le \frac 1 {k!}.$$ Combining this with $$|w_n| \le |w| + 1|w_n| \le |w| + 1$$, $$\sum_{k=K}^n {n \choose k} \frac{|w_n|^k}{n^k} < \sum_{k=K}^n \frac{(|w|+1)^k}{k!} < \frac \varepsilon 3\sum_{k=K}^n {n \choose k} \frac{|w_n|^k}{n^k} < \sum_{k=K}^n \frac{(|w|+1)^k}{k!} < \frac \varepsilon 3$$ Finally, the first sum converges to $$00$$ due to $$w_n \to ww_n \to w$$ and $${n \choose k} n^{-k} \to \frac 1 {k!}{n \choose k} n^{-k} \to \frac 1 {k!}$$. We can choose $$N > KN > K$$ such that it's smaller than $$\varepsilon / 3\varepsilon / 3$$ as soon as $$n > Nn > N$$.