= Solution
The <Cauchy theorem for groups> says that if a prime $p$ divides $|G|$, then $G$ contains an element of order $p$.
Consider
$$
X=\{(x_1,\ldots,x_p)\in G^p:x_1x_2\cdots x_p=1\}.
$$
The first $p-1$ entries determine the last, so $|X|=|G|^{p-1}$, which is divisible by $p$. Cyclic rotation acts on $X$: if the product is one, then
$$
x_2\cdots x_px_1=x_1^{-1}(x_1\cdots x_p)x_1=1.
$$
Every orbit has size one or $p$. The fixed points are exactly the constant tuples $(x,\ldots,x)$ satisfying $x^p=1$. Since $|X|\equiv0\pmod p$, the number of fixed points is divisible by $p$. The identity supplies one, so there is a nonidentity $x$ with $x^p=1$. Its order divides the prime $p$ and is not one, hence is $p$.
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