On the generalized Leibniz rule problem definition
I have to evaluate in $z=0$ the $n$-th derivative with respect $z$ of the product $f(z)\cdot z^k$, where $f(\cdot)$ is a generic smooth function and $k$ is a given integer. I will use the short notation $F^{(n)}(z)$ to denote the $n$-th derivative of the generic function $F(z)$, so what I want is to compute for any $n\in\mathbb{N}$
\begin{equation}
[(f(z)\cdot z^k)^{(n)}]_{z=0}\triangleq \frac{\text{d}^n\left[f(z)\cdot z^k\right]}{\text{d}z^n}\Bigg|_{z=0}
\tag{1}
\end{equation}
my attempt
For the generalized Leibniz rule, holds for any $n$
\begin{equation}(f(z)\cdot z^k)^{(n)}=\sum_{i=0}^n \binom{n}{i}\cdot f^{(n-i)}(z) \cdot \left(z^k\right)^{(i)}\end{equation}
so the problem consist into compute the $i$-th derivative of the power $z^k$. If I'm not wrong,
\begin{equation}
\left(z^k\right)^{(i)}=\begin{cases}
\frac{k!}{(k-i)!}z^{k-i} & \text{if } i\leq k\\
0 & \text{otherwise}
\end{cases}
\end{equation}
This formula says that the index $i$ of the previous summation cannot exceed the value $k$. Anyway, $k$ is an external parameter and can be greater than $n$: in this case the summation stops at the value $n$,
otherwise the summation stops at value $k$.
So I would write
\begin{equation}\begin{aligned}(f(z)\cdot z^k)^{(n)}&=\sum_{i=0}^{\text{min}(n,k)} \binom{n}{i}\cdot f^{(n-i)}(z) \cdot  \frac{k!}{(k-i)!} z^{k-i}\\
&=k!\cdot\sum_{i=0}^{\text{min}(n,k)} \binom{n}{i}\cdot \frac{1}{(k-i)!}\cdot f^{(n-i)}(z) \cdot z^{k-i}\\
\end{aligned}\end{equation}
Now comes the problems. By setting $z=0$ turns out
\begin{equation}\begin{aligned}
%
[(f(z)\cdot z^k)^{(n)}]_{z=0}
&=k!\cdot\sum_{i=0}^{\text{min}(n,k)} \binom{n}{i}\cdot \frac{1}{(k-i)!}\cdot f^{(n-i)}(0) \cdot 0^{k-i}\\
\end{aligned}\end{equation}
from my prospective this expression is quite tricky because of the term $0^{k-i}$. I'm tempted to write
\begin{equation}
0^{k-i}=\begin{cases}1 & \text{if } i=k\\
0 & \text{otherwise}
\end{cases}
\end{equation}
and consequently simplify the last summation as
\begin{equation}\begin{aligned}
%
[(f(z)\cdot z^k)^{(n)}]_{z=0}
&=k!\cdot\sum_{i=0}^{\text{min}(n,k)} \binom{n}{i}\cdot \frac{1}{(k-i)!}\cdot f^{(n-i)}(0) \cdot 0^{k-i}\\
&=\begin{cases}
k!\cdot\binom{n}{k}\cdot \frac{1}{(k-k)!}\cdot f^{(n-k)}(0) \cdot 0^{k-k} & \text{if } n\geq k \\
0 & \text{otherwise}
\end{cases}\\
&=\begin{cases}
\frac{n!}{(n-k)!}\cdot f^{(n-k)}(0) & \text{if } n\geq k \\
0 & \text{otherwise}
\end{cases}\\
\end{aligned}\end{equation}
question
I don't have a precise question about my problem. Essentially I'm doubtful about my derivation because of the undefined power $0^0$.

We consider  $n,k$ non-negative integers. It might be more convenient to consider a slightly simpler representation, namely
\begin{align*}
\color{blue}{\left.\left(f(z)\cdot z^k\right)^{(n)}\right|_{z=0}}
&=\left.\sum_{j=0}^n\binom{n}{j}f^{(n-j)}(z)(z^k)^{(j)}\right|_{z=0}\\
&=\left.\sum_{j=0}^n\binom{n}{j}f^{(n-j)}(z)k(k-1)\cdots(k-j+1)z^{k-j}\right|_{z=0}\tag{1}\\
&=
\begin{cases}
\binom{n}{k}f^{(n-k)}(0)k!&\ \qquad 0\leq k\leq n\\
0&\ \qquad k>n
\end{cases}\\
&\,\,\color{blue}{=
\begin{cases}
\frac{n!}{(n-k)!}f^{(n-k)}(0)&\qquad 0\leq k\leq n\\
0&\qquad k>n
\end{cases}
}
\end{align*}
In the representation (1) we see at a glance that evaluation at $z=0$ is equal to zero if $k\ne j$. This way there is no need to additionally define $0^0:=1$.
