$f$ an entire function, $|f(1/n)|\leq n^{-n}$ for $n\in \mathbb{N}$. Show that $f$ is constant. I'm working on old qualifying exam problems and this one came up. 
Suppose $f$ is an entire function on $\mathbb{C}$, and that $|f(\frac{1}{n})| \leq n^{-n}$ for all $n\in \mathbb{Z}_{>0}$. Prove that $f$ is constant.
My approaches: One obvious thing to do is to try and show that $f$ is bounded and appeal to Liouville's theorem. However, our bounds are for small values of $z$, so this did not yield much success.
One can see that $f(0)=0$ from this condition. Also,
$$
 f'(0) = \lim_{n\to \infty} \frac{f(\frac{1}{n}) - f(0)}{\frac{1}{n}} = \lim_{n\to \infty} n f\left(\frac{1}{n}\right).
$$
Note that
$$
 \left|nf\left(\frac{1}{n}\right)\right| \leq n^{-n+1} \to 0 \text{ as } n\to \infty.
$$
Thus, $f'(0)=0$.
I would like to keep trying to show that $f^{(n)}(0)=0$ and use the fact that $f$ is entire to get that $f=0$ identically on $\mathbb{C}$. But I get stuck because I don't have an estimate for $f'(1/n)$. 
I think what makes this problem harder is that we only have an estimate on $f$ at a countable set, instead of the usual problem where we know some kind of bound on $f$ everywhere.
Any hints are appreciated.
 A: If $f$ is not identically zero then $f$ has a zero of multiplicity $k \ge 0$ at $z=0$, i.e.
$$
f(z) = z^k g(z)
$$
where $g$ is an entire function with $g(0) \ne 0$.
Then $$
\bigl|g(\frac 1n)\bigr| = n^k \, \bigl|f(\frac 1n)\bigr| \le \frac{n^k}{n^n}
$$
for all positive integers $n$. The right-hand side tends to zero for $n \to \infty$
and it follows that $g(0) = 0$, which is a contradiction.
A: Since $f$ is entire, we know
$$f(z) = \sum_{j=0}^\infty a_j z^j \quad \forall z \in \mathbb{C}$$
Using the bound given, we see
$$ \left |f\left (\frac{1}{n}  \right) \right | = \left | \sum_{j=0}^\infty \frac{a_j}{n^j}  \right | \leq \frac{1}{n^n}$$
You've already seen that $a_0=0$ by taking $n\to \infty$, now pull out a factor of $n$ to see
$$\left |f\left (\frac{1}{n}  \right) \right | = \frac{1}{n}\left | \sum_{j=1}^\infty \frac{a_j}{n^{j-1}}  \right | \leq \frac{1}{n^n}\implies  \left|n f\left ( \frac{1}{n} \right) \right | =\left | \sum_{j=1}^\infty \frac{a_j}{n^{j-1}}  \right | \leq \frac{1}{n^{n-1}}$$
i.e. we have that $a_i =0$ for $i \in [0,n]$. Proceed with induction to conclude such a function is identically zero. 
Remark to what you said: You don't need to have the estimate on $f'(1/n)$ since the function is entire and your representation holds for all $z \in \mathbb{C}$.
