Show that the power series $\displaystyle f(z) = 1+z+\frac{z^2}{2!}+ \cdots = \sum_{n=1}^{\infty} \frac{z^n}{n!} $ is equal to $\displaystyle e^z $.

So suppose $\displaystyle w \in \mathbb{C} $. Then $\displaystyle f(z)f(w) =\sum_{n=1}^{\infty} \frac{z^n}{n!} \cdot \sum_{n=1}^{\infty} \frac{w^n}{n!} = f(z+w) $. Also $\displaystyle f(x) = 1+x+\frac{x^2}{2!}+ \cdots = \sum_{n=1}^{\infty} \frac{x^n}{n!} $. And so $\displaystyle f(x) = e^{x} $. And then we need to show that $\displaystyle f(iy) = \cos y + i \sin y $?