Young diagram for exterior powers of standard representation of $S_{n}$

Here is a solution using Pieri's rule:

The representation $\wedge^s V$ has as basis vectors: $$ \{e_{i_1}\wedge \dotsb \wedge e_{i_s}\mid 1\leq i_1<\dotsb <i_s\leq n\}. $$

If we restrict this representation to $S_{n-1}$, then the representation on the subspace spanned by $\{e_{i_1}\wedge \dotsb \wedge e_{i_s}\mid 1\leq i_1<\dotsb <i_s\leq n-1\}$ is just the representation of $S_{n-1}$ on $\wedge^s V_{n-1}$, where $V_{n-1}$ is the subspcace of $V$ consisting of vectors with the last coordinate equal to $0$.

On the other hand, the representation of $S_{n-1}$ on the subspace spanned by $\{e_{i_1}\wedge \dotsb \wedge e_{i_{s-1}}\wedge e_n\mid 1\leq i_1<\dotsb <i_{s-1}\leq n-1\}$ is isomorphic to the representation $\wedge^{s-1}V_{n-1}$ of $S_n$.

Therefore, by induction hypothesis, the restriction of $\wedge^s V$ to $S_{n-1}$ is the sum of the representation corresponding to $(n-s, 1^{s-1})$ and the representation corresponding to $(n-s-1, 1^s)$. It follows from Pieri's rule that $\wedge^s V$ is the representation corresponding to $(n-s, 1^s)$.