Zeros of partial sums of the exponential I am trying to show that if $$f_n(z)=1+z+\frac{z^2}{2!}+...+\frac{z^n}{n!}$$
Then $f_n(z)$ don’t have zeros inside the unitary disk. 
I have tryied to use Rouche’s theorem or use that in the limit the polinomial converges to the exponential, but i dont get hoy to do this. 
 A: You have all the facts you need. Here is an outline of what you need to do:
Use Rouche's theorem with $f_n$ and $z^n\over n!$ to show that if some $f_N$ has a zero within the unit disk, then every $f_n$ with $n\gt N$ has a zero in the unit disk. Now since there are an infinite number of zeros in the unit disk, the set of points mapping to zero under our $f_n$'s has at least one accumulation point. So we have a sequence of pairs, $f_i, p_i$ with $f_i(p_i)=0$ for each i, with a limit point $f, p$ (as the $f_n$'s also converge). For the sake of contradiction, we want to show $f(p)=0$ as we already know that f is the exponential and so has no zeros.
To do this use an epsilon delta argument: let $\epsilon \gt 0$, then find $N_0$ such that $f_n(x)$ is within $\epsilon \over 3$ of f(x) for all x in the unit disk when $n\gt N_0$, then find $\delta$ such that when $|x-x_0|\lt\delta$, then  $|f(x)-f(x_0)| \lt$ $ \epsilon \over 3$. Lastly find $N_1$ so that $n\gt N_1$ implies $|p-p_n| \gt \delta$. Use the triangle inequality to show that when $n>max(N_0,N_1)$ we have $|f(p)-f_n(p_n)|<\epsilon$. This completes the contradiction, as it implies $f(x)=e^x$ has a zero in the unit disk.
