I know this is going to come across as a very strange question, but it's important that I know the answer.
Say I have a degree $k$ polynomial (for my case, I need it to be a complex-valued polynomial with real coefficients, but I think the idea would work even if it were real-valued), $f(z) = a_{0}+a_{1}z+a_{2}z^{2}+\cdots + a_{k-2}z^{k-2}+a_{k-1}z^{k-1}+a_{k}z^{k}$, then what is the formula for the $k-1$st derivative?
Obviously, the first few derivatives are:
$f^{\prime}=a_{1}+2a_{2}z+\cdots (k-2)a_{k-2}z^{k-3}+(k-1)a_{k-1}z^{k-2}+ka_{k}z^{k-1}$
$f^{\prime\prime}=2a_{2} + 3\cdot 2 a_{3}z + 4 \cdot 3 a_{4}z^{2} + \cdots + (k-3)(k-2)a_{k-2}z^{k-4}+(k-2)(k-1)z^{k-3}+(k-1)ka_{k}z^{k-2}$
$f^{\prime\prime\prime}=3\cdot 2 a_{3} + 4(3)(2) a_{4}z + \cdots + (k-4)(k-3)(k-2)a_{k-2}z^{k-5}+(k-3)(k-2)(k-1)z^{k-4}+(k-2)(k-1)ka_{k}z^{k-3}$
I'm thinking the $k-1$st derivative would definitely contain only two terms, a constant term and a linear term, and the coefficient of the linear term would be $(k-(k-1))(k-(k-2))\cdots (k-2)(k-1)k$, but I can't figure out the pattern for what the constant term should be.
Could somebody please help me?!
The $(k-1)$st derivative of $a_0 + a_1 z + \cdots + a_{k-1}z^{k-1} + a_k z^k$ is given by $$(k-1)! a_{k-1} + k! a_k z.$$ The pattern might be more obvious if we write that as $$(k-1)! a_{k-1} + \frac{k!}{1!} a_k z.$$ To prove things like this, you should note that differentiation is linear, so you only need to consider monomials. Then, as it's clear that monomials with degree less than $k-1$ are annihilated, you only need to consider the last two terms.