alternating and symmetric powers of the standard representation of the symmetric group Let $n \geq 7$ and $V = \mathbb{C}^n$ be the standard representation for $S_{n+1}$, the symmetric group of cardinal $(n+1)!$ 
Let $k$ be an integer such that $2 \leq k \leq n$. Is it true or false that $\bigwedge^k V$ does not appear as an irreducible sub-representation of $\mathrm{Sym}^k V$? I am looking for a reference which would either prove or give counter-examples to this.
Thanks in advance!
EDIT : Is there an online software (analogous to LIE) that would compute the decompositions into irreducible representations of the tensor product of two ireps for the symmetric group?
 A: Here is another attempt. It follows from Enumerative Combinatorics, vol. 2, Exercise 7.73,
that the Frobenius characteristic of the representation
$\mathrm{Sym}^k V$ is the coefficient of $q^k$ in the power series
  $$ (1-q)\sum_{\lambda\vdash n+1}s_\lambda(1,q,q^2,\dots)s_\lambda. $$
(The factor $1-q$ appears because Exercise 7.73 deals with the
symmetric algebra of the defining representation of degree $n$. The
defining representation is the sum of $V$ and the trivial
representation.) By Corollary 7.21.3,
  $$ s_\lambda(1,q,q^2,\dots) = q^{b(\lambda)} +\ \mbox{higher order
  terms}, $$
where $b(\lambda) = \sum (i-1)\lambda_i$. Hence the least $k$ for
which an irrep indexed by $\lambda$ appears in $\mathrm{Sym}^k V$ is
$b(\lambda)$. For $\bigwedge^k V$ we have $b(\lambda) =b(n-k+1,1^k) =
{k+1\choose 2}>k$ for $k>1$.
A: 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)
