 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 $s_{i}w=-w$ for every adjacent transposition $s_i$. These generators determine the representation, and the sign representation sends every one of them to $-1$. ConsequentlyAt $n=1$ there are no adjacent generators and both modules are the trivial representation.