Was reading Ireland-Rosen intro to number theory and ran into an expression on page 75 I couldn't make sense of.
They assert that (it is easy to see that) the coefficient of $z^\frac{p-1}{2}$ in the power series expansion of $e^{(2k-1)z}-e^{z(p-(2k-1))}$ is $4k-p-2$, which is clearly just the difference $(2k-1)-(p-(2k-1))$.
But I don't know how they got there, I expect something much more ugly and hard to work with looking like $\frac{(2k-1)^\frac{p-1}{2}-(p-(2k-1))^\frac{p-1}{2}}{(\frac{p-1}{2})!}$ which I then try to binomial expand and cancel terms but I don't see how they got to $4k-p-2$.
Picture included for fun times.
The factor $4k-p-2$ is the coefficient of $z^1$, not of $z^{(p-1)/2}$, in the factor $$ \begin{split} e^{(2k-1)z} - e^{(p-(2k-1))z} & = \bigl( 1 + (2k-1)z + O(z^2) \bigr) - \bigl( 1 + (p-2k+1)z + O(z^2) \bigr) \\ & = (4k-p-2) z + O(z^2) . \end{split} $$ It's only when you take the product that you get the coefficient of $z^{(p-1)/2}$ in the whole thing; this is because $$ \bigl( a_1 z + O(z^2) \bigr) \bigl( a_2 z + O(z^2) \bigr) \cdots \bigl( a_n z + O(z^2) \bigr) = a_1 a_2 \cdots a_n z^{n} + O(z^{n+1}) . $$