Why does the polynomial $x^{n+1}-n!$ have $n+1$ distinct roots?

Question

Why does the polynomial $x^{n+1}-n!$ have $n+1$ distinct roots in the field of complex numbers?

Answer 1

As a polynomial of degree $n+1$, $x^{n+1}-n!$ has $n+1$ complex roots.

To show that roots are distinct we check the derivative of $x^{n+1}-n!$.

Note that if a function f(x) has a root of multiplicity more than $1$ then its derivative will have the same root with multiplicity at least $1$

$$ \frac {d}{dx} ( x^{n+1}-n!)=(n+1)x^n$$ does not have any non zero root and zero is not a root of $x^{n+1}-n!$

Thus $x^{n+1}-n!$ does not have any roots of multiplicity more than $1$

Answer 2

$$ \left(\frac{x}{(n!)^{1/(n+1)}}\right)^{n+1}-1=0 $$

the $n+1$ unity roots.

Answer 3

This polynomial has derivative $(n+1)x^n$, the only root of which is $0$. Since $0$ is not a root of $x^{n+1}-n!$, all its complex roots have multiplicity $1$, i.e. are all distinct.

Answer 4

$x^n-a$ has allways $n$ different roots if $a>0$. They are $a^{1/n}e^{2\pi i k/n}$, $k$ from $1$ to $n$.