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I need to show that the product of all the $n$th roots of unity is $(−1)^{n+1}$.

Is there a way to do this by induction? If there is, I can't seem to figure it out. Are there other, perhaps more efficient, methods of proof?

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The problem with attempting to do it "by induction" is that the $n$th roots of unity are not related to the $(n+1)$th roots of unity (except for the single one that they have in common, $1$. – Arturo Magidin Mar 2 '12 at 4:07
If you're going to do an induction argument, it should be an induction on the number of (possibly repeated) prime divisors of $n$. I think this should actually be possible, with Wilson's Theorem stepping in as a base case. But, honestly, just about anything else seems all the answers given below. – Pete L. Clark Mar 2 '12 at 4:28
up vote 9 down vote accepted

The $n^{th}$ roots of unity are precisely the roots of the polynomial $$ x^n - 1 $$ Now use the fact that the product of the roots of any polynomial is $(-1)^n$ times the constant coefficient.

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Is this fact that the product of the roots of any polynomial is $(−1)^n$ times the constant coefficient true for any polynomial? Can you elaborate on that? – Dominick Gerard Mar 2 '12 at 4:42
@DominickGerard: Consider writing the polynomial as a product of linear factors, then multiplying it out again. What will the constant term be? Example: $(z-z_1)(z-z_2) = z^2 - (z_1 + z_2) z + z_1 z_2$. – Antonio Vargas Mar 2 '12 at 4:58
@DominickGerard Yes, it holds for any polynomial with coefficients in an integral domain (including $\mathbb{Z}$ or any field): – dls Mar 2 '12 at 5:01

Note that if $\zeta$ is an $n$th root of unity, then so is $\frac{1}{\zeta}$. Moreover, the only roots of unity for which $\zeta=\frac{1}{\zeta}$ are $\zeta=1$ and $\zeta=-1$ (since $\zeta=\frac{1}{\zeta}$ implies $\zeta^2=1$, hence $\zeta=1$ or $\zeta=-1$).

So, if we take the $n$th roots of unity, then we can pair off all of them except for $1$ and perhaps $-1$ (if $n$ is even), into pairs of the form $\{\zeta,\frac{1}{\zeta}\}$. The product of these two equals $1$; so when you do the product of all roots, you end up with a bunch of pairs that multiply to $1$, and you have $1$; and if $n$ is even, then you also have $-1$ leftover. Which means the product of all of them is...

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Well, you said "efficient" and "elementary number theory", and that triggered this in my mind: the result is a special case of

(Wilson's Theorem in a Finite Abelian Group): Let $(G,\cdot)$ be a finite abelian group, and let $P = \prod_{x \in G} x$. Then $P = 1$ (the identity element) unless $G$ has exactly one element $t$ of order $2$, in which case $P = t$.

This is something that others (in particular, Bill Dubuque) have talked about here and elsewhere before. I include a statement and proof of this in my notes/prebook on elementary(ish...) number theory: it is at the end of Appendix B.

Added: Not only is this a special case of a "generalized Wilson theorem", it's a special case that includes the classical Wilson Theorem! Since the unit group of $(\mathbb{Z}/p\mathbb{Z})^{\times}$ is cyclic of order $p-1$, computing $(p-1)! \pmod{p}$ is essentially the same as multiplying together all the $(p-1)$st roots of unity.

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+1 See these posts for more on the group-theoretic Wilson's theorem and involutions. – Bill Dubuque Mar 2 '12 at 4:13

You can prove it with de Moivre, on the basis that the roots are all of the form $e^{2\pi i k/n}$, so their product is $e^{2\pi i/n\sum{k}} = e^{\pi i(n-1)} = (-1)^{n-1} = (-1)^{n+1}$.

Another proof idea would be to notice that the roots of unity come in complex conjugate pairs, which is evident in the symmetry of the spacing of the roots in the complex plain. So the product of each pair is of course $\alpha \bar\alpha = |\alpha| = 1$ since all the roots fall on the unit circle. The only exceptions are $1,-1$. $1$ is always a root and won't change the product, and $-1$ is only a root when $n$ is even, so we conclude that the product is exactly $(-1)^{n+1}$.

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