These are zeros of $g'(x)$, where $$g(x) :=\frac{(x+n_1)\cdots(x+n_k)}{x^{k-1}}.$$ Or, switching to $y:=\frac1x$, they correspond to zeroes of $f'(y)$, where $$f(y) := \frac{(1+n_1y)\cdots(1+n_ky)}y.$$
Max Alekseyev
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