The strategy is to observe that the order of an element equals the order of the cyclic subgroup it generates, then apply [Lagrange's Theorem](/theorems/782). **Step 1: Relate element order to subgroup order.** Let $g \in G$ with $o(g) = n$. The cyclic subgroup $\langle g \rangle = \{e, g, g^2, \ldots, g^{n-1}\}$ has order $|\langle g \rangle| = n = o(g)$. **Step 2: Apply Lagrange.** Since $\langle g \rangle \leq G$, [Lagrange's Theorem](/theorems/782) gives $|\langle g \rangle| \mid |G|$, i.e., $o(g) \mid |G|$. **Step 3: Derive the exponent identity.** Write $|G| = k \cdot o(g)$ for some positive integer $k$. Then: \begin{align*} g^{|G|} = g^{k \cdot o(g)} = (g^{o(g)})^k = e^k = e. \end{align*}