$\mathbf{Question}:$ For a function $f: \mathbb{C} \to \mathbb{C}$ and $n \geq 1$, let $f^{(n)}$ denote the $n$th derivative of $f$ and $f^{(0)}=f$. Let $f$ be an entire function such that for some $n\geq 1$, $f^{(n)}(\frac{1}{k})=0$, $\ \forall k\in \mathbb{N}$. Show that $f$ is a polynomial.
$\mathbf{Attempt}:$ $f$ being an entire function, we can express $f$ as $f(z)= \displaystyle\sum_{t=0}^\infty a_tz^t $.
We get $f^{(n)}(z)=\displaystyle\sum_{t=n}^\infty t(t-1)...(t-n+1)a_t z^{t-n}$
Firstly, we have $\displaystyle\lim_{k \to \infty}f^{(n)}\bigg(\frac{1}{k}\bigg)=n!a_n=0 \implies a_n=0$ ($k$ is natural)
Now, for $f^{(n)}(z)=\displaystyle\sum_{t=n+1}^\infty t(t-1)...(t-n+1)a_t z^{t-n}= z\displaystyle\sum_{t=n+1}^\infty t(t-1)...(t-n+1)a_tz^{t-n-1} \ $, and for $z=1/k$, $f^{(n)}(\frac{1}{k})=0 \implies \displaystyle\sum_{t=n+1}^\infty t(t-1)...(t-n+1)a_t\bigg(\frac{1}{k}\bigg)^{t-n-1}=0 $. Again, taking $k \to \infty $, we get $a_{n+1}=0$.
Let $a_{n+m}=0$. Then $f^{(n)}(z)=\displaystyle\sum_{t=n+m+1}^\infty t(t-1)...(t-n+1)a_t z^{t-n}=z^{m+1}\displaystyle\sum_{t=n+m+1}^\infty t(t-1)...(t-n+1)a_t z^{t-n-m-1}$
Now, $(\frac{1}{k})^{m+1}\displaystyle\sum_{t=n+m+1}^\infty t(t-1)...(t-n+1)a_t (1/k)^{t-n-m-1}=0 \implies \displaystyle\sum_{t=n+m+1}^\infty t(t-1)...(t-n+1)a_t (1/k)^{t-n-m-1}=0 $. Taking $\displaystyle\lim_{k \to ∞}\sum_{t=n+m+1}^\infty t(t-1)...(t-n+1)a_t (1/k)^{t-n-m-1}=0$, we are left with $a_{n+m+1}=0$.
Thus we conclude that $a_{n+p}=0$ for all $p \in \mathbb{N}$
Now, $f^{(n)}(z)=0$, which gives us $f(z)=c_0+c_1z+...c_{n-1}z^{n-1}$.
Is this method correct?
Kindly $\mathbf{Verify}$
PS: I am aware of the duplicates, I just want to get my method checked without me looking at the previously answered questions.
Write $g(z) = f^{(n)}(z)$ where $n$ has the property you described. The zeros of a nonconstant analytic function are isolated. Consider the sequence $a_k = \frac{1}{k}$, then since $g$ is analytic we find that $$g(0) = \lim_{k\to \infty} g(a_k) = \lim_{k\to \infty }0 = 0$$ Thus $0$ is not an isolated zero of $g$. We conclude that since $g$ is analytic, then it is constant and in particular $g= 0$. Now the power series representation in a domain about a point is unique. From here we can work backwards using antiderivatives to determine that $g(z) = c_o + c_1z + \ldots + c_{n-1}z^{n-1}$.
Edit: Sorry, just saw your request to verify. You should be more clear what is going on with your first statement. I think you are trying to show that $f^{(n)}(0) = 0$ and thus you find that the constant term is zero? For the second part, I think it would be clearer if you said "$f^{(n)}(0) = 0$ and we have shown that the constant term is zero.... we can write the expansion as $f^{(n)}(z) = z\sum_{k=n+1}^\infty \alpha_k z^{k-n-1}$. Well $f^{(n)}(\frac{1}{k}) = (\frac{1}{k})\sum_{k=n+1}^\infty \alpha_k z^{k-n-1} = 0$. Thus $\sum_{k=n+1}^\infty \alpha_k z^{k-n-1} = 0$ whenever $z\in\{0\}\cup\{\frac{1}{k}:k\in\mathbb{N}\}$". I think this is what you are doing and then say iterate?