Showing that the product of all $n$th Roots of Unity is $(-1)^{n+1}$ 
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?
 A: 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.
A: 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 $(-1)^{n+1}$.
A: 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.
A: 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...
