Schur multiplier of finite group is finite and exponent of Schur multiplier divides group order

Statement
Let $$G$$ be a finite group. Then the Schur multiplier $$M(G)$$ of $$G$$ is finite, and its exponent divides the order of $$G$$.

Proof
Let $$n$$ be the order of $$G$$.

We show that any 2-cocycle for the action of $$G$$ on $$\mathbb{C}^*$$ is cohomologous (viz, a coboundary times) a 2-cocycle with values in the $$n^{th}$$ roots of unity. To see this let $$\alpha$$ be a 2-cocycle $$\alpha: G \times G \to \mathbb{C}^*$$. Then:

$$\alpha(x,y)\alpha(xy,z) = \alpha(x,yz)\alpha(y,z)$$

We now take the product over all $$x \in G$$. We get:

$$\prod_{x \in G} \alpha(x,y)\alpha(xy,z) = \prod_{x \in G} \alpha(x,yz) \alpha(y,z)$$

Set $$t(y) = \prod_{x \in G} \alpha(x,y)$$. The above then simplifies to:

$$t(y)t(z) = t(yz)\alpha(y,z)^n$$

For any $$x \in G$$ choose $$g(x) \in \mathbb{C}^*$$ such that $$g(x)^n = t(x)^{-1}$$. Then the coboundary $$\partial g$$ is the map $$(x,y) \mapsto g(xy)/g(x)g(y)$$ and we can easily see that $$\beta = (\partial g) \alpha$$ is a 2-cocycle such that $$\beta(x,y)^n = 1$$ for all $$x,y \in G$$. This completes the proof.