 Solution

The symmetric group acts transitively on the Young tableaux of a fixed shape. Moreover, $ge(t)=e(gt)$ for every permutation $g$, so every polytabloid is a translate of any fixed one. Hence the Specht module is a cyclic module generated by $e(t)$.

It remains to see that $e(t)\ne 0$. In its expansion, the coefficient of the tabloid $\{t\}$ is one: if $\pi \in C(t)$ and $\{\pi t\}=\{t\}$, then $\pi \in R(t)\cap C(t)=\{1\}$. Thus the generator, and therefore the module, is nonzero.