Let $W = \mathbb C^n$, with the obvious representation of $S_n$.

**Lemma** Let $\lambda$ be a partition of $n$. The multiplicity of the irreducible representation $V_\lambda$ inside $W^{\otimes k}$ is given by the number of ways of obtaining the partition $\lambda$ by starting with the partition $[n]$, and then alternating between removing a box from the Ferrers diagram and adding a box to the Ferrers diagram $k$ times.

**Proof** Induction on $k$, the base case $k=0$ being trivial. We have
$$ W^{\otimes (k+1)} = W^{\otimes k} \otimes \mathrm{Ind}_{S_{n-1}}^{S_n} \mathbb C = \mathrm{Ind}_{S_{n-1}}^{S_n} \mathrm{Res}^{S_n}_{S_{n-1}} W^{\otimes k},$$
see e.g. Exercise 3.16 of Fulton-Harris. Thus the result follows from Pieri's rule. **QED**

Let $V = W/\mathbb C$ be the irreducible representation corresponding to the partition $[n-1,1]$. Then $\wedge^k V$ is also irreducible, with corresponding partition $[n-k,1,1,\ldots,1]$, for $k\leq n-1$. By the preceding lemma, it is easy to see that the multiplicity of $\wedge^k V$ inside $W^{\otimes k}$ is exactly $1$ for $k \leq n-1$.

It follows in particular that $\wedge^k V$ occurs with multiplicity $0$ inside $\mathrm{Sym}^k(W)$ for $k=2,\ldots,n-1$. Indeed $\wedge^k(V) \subset \wedge^k(W)$ and $\wedge^k(W) \cap \mathrm{Sym}^k(W) = \{0\}$ for $k \geq 2$, where the intersection is taken inside of $W^{\otimes k}$. In particular this representation doesn't occur inside $\mathrm{Sym}^k(V)$, either.

Also, here is some SAGE code that will compute things for you.

```
s = SymmetricFunctions(QQ).schur()
V = s([n-1,1])
ext = s([1]*k)
sym = s([k])
print ext.inner_plethysm(V) ## decomposition of \wedge^k(V)
print sym.inner_plethysm(V) ## decomposition of Sym^k(V)
```