TITLE: Entire function constant, where $f(z)=f(z+1)$ and $|f(z)|< e^{|z|}$.
QUESTION [7 upvotes]: I came across this old exam problem.

Suppose $f(z)$ is entire and $|f(z)|< e^{|z|}$, and also $f(z)=f(z+1)$. Show $f(z)$ is a constant.

I am able to show the singularity at infinity is not a pole.   But I can't rule out it being an essential singularity.

REPLY [4 votes]: If you don't want to use the machinery of the extension theorem, define the function 
$$g(z):= \frac{f(z)-f(0)}{e^{-i2\pi z}-e^{i2\pi z}}.$$
Where $|f(z)| \leq e^{C |z|}$
(the denominator is a constant times $\sin$.)
Then it's easy to check that g(z) is entire given the periodicity conditions on $f$. Then we can show that $g(z)$ is bounded on the strips $0 \leq \Re(z) \leq 1$ (and therefore everywhere). Let $z=a + bi$ (where $a$ and $b$ are real. Then
$$|g(z)| = |g(a+ bi)| \leq \frac{e^{C|b|}+1}{e^{2 \pi b}-1}$$
when $b>0$ and $$ |g(z)| \leq \frac{e^{C|n|}+1}{e^{-2\pi b}-1}$$
when $b<0$
by using the reverse triangle inequality both ways in the denominator.
Then if $C<2 \pi$, the denominator will dominate for $b$ large so $g(z)$ is bounded at infinity. It is clearly bounded away from infinity. 
By Liouville, $g(z) = A$ for some constant, or
$$f(z) = f(0) + A \sin(2\pi z)$$ 
From here it's not hard to show $A=0$