<s>[INST] $P(z)=(z^n-1)$ has roots $1, a_1,a_2,...,a_{n-1} $. Prove that $(1-a_1)(1-a_2)...(1-a_{n-1})=n$ $P(z)=(z^n-1)$ has roots $1, a_1,a_2,...,a_{n-1} $. Prove that $(1-a_1)(1-a_2)...(1-a_{n-1})=n$
No idea how to prove this. Would really appreciate any help, hints...
Thanks in advance :)
 [/INST] Since
$P(z)= z^n - 1, \tag 1$
we have
$P(z) = (z - 1)(z^{n - 1} + z^{n - 2} + \ldots + z + 1) = (z - 1)Q(z), \tag 2$
where
$Q(z) = z^{n - 1} + z^{n - 2} + \ldots + z + 1 = \sum_0^{n - 1} z^i \tag 3$
it then follows that if $1 \ne b \in \Bbb C$ satisfies
$P(b) = (b - 1)(b^{n - 1} + b^{n - 2} + \ldots + b + 1) = (b - 1)Q(b) = 0, \tag 4$
we must have
$b^{n - 1} + b^{n - 2} + \ldots + b + 1 = \sum_0^{n - 1} b^i = Q(b) = 0; \tag 5$
thus any root $b$ of $P(z)$ other than $1$ must satisfy $Q(z)$; now $\deg Q(z) = n - 1$, so it has at most $n - 1$ zeroes, so if we assume the list of roots of $P(z)$,
$1, a_1, a_2, \ldots, a_{n - 1}, \tag 6$
is irredundant, that is, does not repeat itself (as seems to be the intention of our OP kjhg), then the zeroes of $Q(z)$ are the $a_i$, $1 \le i \le n - 1$. Thus we may write
$Q(z) = \displaystyle \prod_1^{n - 1} (z - a_i); \tag 7$
comparing (3) and (7), we see that
$\displaystyle \prod_1^{n - 1} (z - a_i) = z^{n - 1} + z^{n - 2} + \ldots + z + 1 = \sum_0^{n - 1} z^i;\tag 8$
taking $z = 1$ in (8), we reach
$\displaystyle \prod_1^{n - 1} (1 - a_i) = 1^{n - 1} + 1^{n - 2} + \ldots + 1^1 + 1 = n, \tag 8$
the desired result.
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