$$f(z)= 1 + \frac{1}{z} + \frac{1}{2!z^2} + ... + \frac{1}{n!z^n}$$
I can't seem to apply Rouche's Theorem here.
Another idea would be to use perhaps Big Picard's theorem, since f(z) will converge to $e^{\large \frac{1}{z}}$, I think. And so it will have an essential singularity at z=0.
But Big Picard allows for one exceptional point. What if the exceptional point is...0? Then I am stuck.
Any ideas are welcome.
Thanks,
Using Rouche's theorem we prove that for arbitrary large $R$, there exists $n$ such that $$f(1/z)=1+\sum_{k=1}^n z^k/k!\ne 0$$ inside $|z|=R, \,$ that is, $f(1/z)$ has all roots outside $|z|=R\,(=1/r)$.
It is all we shall prove.
Since the series $1+\sum_{k=1}^\infty z^k/k!$ converges to $e^z$ uniformly on $|z|\le R $, $$\left|-\sum_{k=n+1}^\infty z^k/k!\right|<\varepsilon \quad (|z|\le R)$$ for sufficiently large $n$.
Take $\varepsilon =e^{-R}$, then since $\min_{|z|=R} |e^z|=e^{-R}$ we have $$ \left|-\sum_{k=n+1}^\infty z^k/k!\right|< |e^z|$$ on $|z|=R$.
Therefore by Rouche's theorem $e^z$ and $f(1/z)=e^z+\left(-\sum_{k=n+1}^\infty z^k/k!\right)$ have the same number of zeros, that is, $f(1/z)$ has no zeros inside $|z|=R$.