Degree of a polynomial as smooth map - clarification

Question

I would like to a clarification of the following question. I cannot understand the last three lines, where we conclude that the exponents $n_1=n_2=...n_k=1$, which then concludes our proof. Cannot understand why the condition of the jacobian implies the one on the $n_s$. Any help is appreciated! Thanks!

Answer 1

Since $q$ is a regular value, then the derivative of the polynomial at every root is different from $0$.

Note that the only possibility to have the derivative different from $0$ at every $x_j$ for all $j=1,...,k$ is that $n_1=...=n_k=1$, since in this case the derivative does not equal $0$.

By the condition that the sum of all the $n$'s is equal to $d$, you easily get the result recalling that for holomorphic proper functions $deg(F)=\#F^{-1}(q)$.