TITLE: alternating and symmetric powers of the standard representation of the symmetric group
QUESTION [9 upvotes]: 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?

REPLY [6 votes]: 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 $k>1$.