= Solution
The single-column <Young diagram> has exactly one <standard Young tableau>, so $V^{(1^n)}$ has dimension one. Consecutive labels are in one column, and the <Young seminormal form> gives $s_iw=-w$ for every adjacent <transposition> $s_i$. These generators determine the representation, and the <sign representation> sends every one of them to $-1$. Consequently
$$
\boxed{V^{(1^n)}\cong\operatorname{sgn}.}
$$
At $n=1$ there are no adjacent generators and both modules are the <trivial representation>.
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