Consider this infinite product: $$\prod _{n=1}^\infty \Bigl( 1+ \frac z n\Bigl) e^{-\frac z n}\ .$$ I must show that it converges to an entire function $f$, and determine the zeros of $f$. It's immediate that $f$ will vanish only in the points $\{-n:n\in \mathbb Z^+\}$, since $f(z)=0$ if and only if one of the factors vanishes in $z$. The problem is to prove that the product actually converges to an entire function, i.e. that it converges uniformly over any compact set of $\mathbb C$.
First off, it's necessary to prove pointwise convergence: as $n\to \infty$ we have $$\Bigl( 1+ \frac z n\Bigl) e^{-\frac z n}-1 \sim \Bigl( 1+ \frac z n\Bigl)\Bigl(1-\frac z n+\frac {z^2} {2n^2}\Bigl)-1=-\frac {z^2} {n^2} +\frac {z^2} {2n^2}=\frac {z^2} {n^2}\ .$$ This proves pointwise convergence for every $z$ (I didn't write the $o\bigl(\frac {z^2} {n^2}\bigl)$ that clearly converge too).
In order to prove uniform convergence over a generic compact set, say $A\subset \mathbb C$, let's define $M=\sup _A|z|$ and $m=\inf_A |z|$. We have $$\Bigl| \Bigl( 1+ \frac z n\Bigl) e^{-\frac z n} -1 \Big|\le \Bigl( 1+ \frac M n\Bigl) e^{-\frac m n} -1 \le \Bigl( 1+ \frac M n\Bigl) \Bigl(1-\frac m n + \frac {m^2} {2n^2} \Bigl) -1\ .$$ However here it remains the term $\frac {M-m} n$, so we can't say that the sum converges. How do I avoid this problem? The chain of inequalities above is the only one that seems possible to me, so clearly I am missing something. Thanks in advance for your help.
Your inequalities are not valid. You are treating complex numbers the same way you treat real numbers. You cannot use $m$ to bound $e^{-\frac z n}$.
Actually your argument for pointiwise convergence almost gives uniform convergence on compact sets: The Taylor expansion that you have used actually shows that $|1-(1+\frac z n) e^{-\frac z n}| \leq \frac C {n^{2}}$ for some constant $C$ whenever $|z|$ is bounded. So M-test finishes the proof.