Glasser's derivation of trinomial equation $x^N − x + t = 0$ from https://arxiv.org/pdf/math/9411224.pdf
I am trying to understand this in more detail
$
x^N - x + t=0 \,\!
$
In particular, the quintic equation can be reduced to this form by the use of Tschirnhaus transformations as shown above. Let $x = \zeta^{-\frac{1}{N-1}}\,$, the general form becomes:
$
\zeta = e^{2\pi i} + t\phi(\zeta) \,\!
$
where
$
\phi(\zeta) = \zeta^{\frac{N}{N-1}} \,\!
$
A formula due to Joseph Louis Lagrange Lagrange  states that for any analytic function $f \,$, in the neighborhood of a root of the transformed general equation in terms of $\zeta \,$, above may be expressed as an  infinite series :


*$
f(\zeta) = f(e^{2\pi {\mathrm{i}}}) + \sum^\infty_{n=1} \frac{t^n}{n!}\frac{d^{n-1}}{da^{n-1}}[f'(a)|\phi(a)|^n]_{a = e^{2\pi {\mathrm{i}}}}
$
If we let $f(\zeta) = \zeta^{-\frac{1}{N-1}}\,$ in this formula, we can come up with the root:


*$
x_k = e^{-\frac{2k\pi {\rm{i}}}{N -1}} - \frac{t}{N-1}\sum^\infty_{n=0}\frac{(te^{\frac{2k\pi {\rm{i}}}{N-1}})^n}{\Gamma(n + 2)}\cdot \frac{\Gamma\left(\frac{Nn}{N-1} + 1\right)}{\Gamma\left(\frac{n}{N-1} + 1\right)} $
$ k=1,2, 3, \dots , N-1 \,$
I am stuck getting from the penultimate step to the final step.  What does this mean:
$\frac{d^{n-1}}{da^{n-1}}[f'(a)|\phi(a)|^n]_{a = e^{2\pi {\mathrm{i}}}} $
Why is phi being introduced again. What are the first few derivatives such as for $n=1,2$.
 A: I was able to recreate the author's steps
(replacing zeta with x because easier to write)
Step 1, the derivative  $f'(x)$ is taken
$f(x) = x^{-\frac{1}{N-1}}\,$
which gives $-1/(N-1) x^{-N/(N-1)}$
Then this is multiplied by  $
\phi^{n+1}(x) = x^{\frac{(n+1)N}{N-1}} \,\!
$
note that since $n=n+1$, it means the sum is slightly different but the same but easier to follow :
$
f(\zeta) = f(e^{2\pi {\mathrm{i}}}) + \sum^\infty_{n=0} \frac{t^{n+1}}{(n+1)!}\frac{d^{n}}{da^{n}}[f'(a)|\phi(a)|^{n+1}]_{a = e^{2\pi {\mathrm{i}}}}
$
which gives $-1/(N-1) x^{Nn/(N-1)}$
$\frac{d^{n}}{da^{n}}[-1/(N-1) x^{Nn/(N-1)}]_{x = e^{2\pi {\mathrm{i}}}} $
which is equal to
$-x^{n/(-1 + N)}/(N-1) * Pochhammer[(-1 + n + N)/(-1 + N), n] $
https://www.wolframalpha.com/input/?i=d%5En%2Fdx%5En+x%5E%28N*n%2F%28N-1%29%29
The Pochhammer evaluates to the gamma function , just like the author
https://www.wolframalpha.com/input/?i=Pochhammer%5B%28-1+%2B+n+%2B+N%29%2F%28-1+%2B+N%29%2C+n%5D+
letting $x=e^{2i\pi}$ in the above expression completes the derivation assuming $k=1$ for the first root
