The admissible-tableau count equals the Littlewood--Richardson coefficient

First identity. Since $s_{\mu }$ is symmetric by [F1] and $w$ acts on monomials by $w(x^{\alpha })=x^{w\alpha }$, for every $w\in S_r$ one has $x^{w(\lambda +{\rho }_r)}{s}_{\mu }=w\left(x^{\lambda +{\rho }_r}{s}_{\mu }\right)={\sum }_T{x}^{w(\lambda +{\rho }_r+\mathrm{wt}(T))}\mathrm{.}$ Multiplying by $\mathrm{sgn}(w)$, summing over $w$, and using $a_{\eta }={\sum }_w\mathrm{sgn}(w)x^{w\eta }$ gives $a_{\lambda +{\rho }_r}{s}_{\mu }={\sum }_T{a}_{\lambda +{\rho }_r+\mathrm{wt}(T)}$.

The bad guys cancel. Call $T$ bad if $\lambda +\mathrm{wt}(T_{\ge j})$ fails to be a partition for some $j\ge 1$; equivalently ${\lambda }_k+\mathrm{wt}(T_{\ge j}{)}_k<{\lambda }_{k+1}+\mathrm{wt}(T_{\ge j}{)}_{k+1}$ for some pair $(k,j)$. Among the pairs $(k,j)$ with $j$ maximal and then $k$ minimal, one has: $\lambda +\mathrm{wt}(T_{>j})$ is a partition (by maximality of $j$), the difference $\mathrm{wt}(T_{\ge j}{)}_k-\mathrm{wt}(T_{\ge j}{)}_{k+1}$ changes by at most one when passing from $T_{>j}$ to $T_{\ge j}$, and hence column $j$ contains a $k+1$ and no $k$, with ${\lambda }_k+\mathrm{wt}(T_{\ge j}{)}_k+1={\lambda }_{k+1}+\mathrm{wt}(T_{\ge j}{)}_{k+1}\mathrm{.}$ Let $T^{\ast }$ be obtained from $T$ by applying the Bender--Knuth involution ${\sigma }_k$ to the subtableau $T_{<j}$ and leaving the rest unchanged. This is well defined and involutive: by the last sentence of [F2] the free cells of $T_{<j}$ are the free cells of $T$ lying in columns $<j$, so the modification swaps the counts of free $k$'s and free $k+1$'s in each row of $T_{<j}$; row weak increase within $T_{<j}$ follows from the Bender--Knuth lemma. Across its boundary, only a $k$ changed to $k+1$ could cause a problem. But column $j$ contains no $k$, so a boundary neighbour in that column which was at least $k$ is at least $k+1$. Hence it remains at least the changed entry (Bender--Knuth involutions permute the weights of semistandard tableaux); column strictness is preserved because each column changes in at most one cell, as in the proof of Bender--Knuth involutions permute the weights of semistandard tableaux. Moreover $(T^{\ast }{)}_{\ge j}=T_{\ge j}$, so $T^{\ast }$ is bad again, and the same pair $(k,j)$ is selected for $T^{\ast }$: the violation tests at all levels $j^{\mathrm{\prime }}\ge j$ are unchanged, so $j$ is still maximal; and the test at level $j$ is unchanged, so $k$ is still minimal. Hence applying ${\sigma }_k$ to $T_{<j}^{\ast }={\sigma }_k(T_{<j})$ returns $T$, and $T\mapsto T^{\ast }$ is an involution of the set of bad guys.

Cancellation. By [F2], $\mathrm{wt}(T_{<j}^{\ast })=s_k\mathrm{wt}(T_{<j})$ and $\mathrm{wt}(T_{\ge j}^{\ast })=\mathrm{wt}(T_{\ge j})$. The equality in step 1.2 says that $s_k$ fixes $\lambda +\mathrm{wt}(T_{\ge j})+{\rho }_r$, so $s_k(\lambda +\mathrm{wt}(T)+{\rho }_r)=\lambda +\mathrm{wt}(T^{\ast })+{\rho }_r$. Since $a_{w\eta }=\mathrm{sgn}(w)a_{\eta }$ [F2] and the transposition $s_k$ is odd, $a_{\lambda +\mathrm{wt}(T^{\ast })+{\rho }_r}=a_{s_k(\lambda +\mathrm{wt}(T)+{\rho }_r)}=-a_{\lambda +\mathrm{wt}(T)+{\rho }_r}$, so the paired terms cancel; if $T=T^{\ast }$, its alternant equals its negative and is zero over $\mathbb{Z}$ in the sum of step 1.1. The bad guys therefore contribute $0$, and the surviving tableaux are exactly the admissible ones, proving the first identity of (i).

Second identity. For admissible $T$ the vector $\lambda +\mathrm{wt}(T)$ is a partition with at most $r$ parts (take $j=1$), so by the bialternant formula [F1] $a_{\lambda +\mathrm{wt}(T)+{\rho }_r}=a_{{\rho }_r}{s}_{\lambda +\mathrm{wt}(T)}$. Substituting into step 2.1 and cancelling the nonzero polynomial $a_{{\rho }_r}$ gives $s_{\lambda }{s}_{\mu }={\sum }_{T\text{ admissible}}{s}_{\lambda +\mathrm{wt}(T)}$.

The admissible-tableau count is the LR coefficient. By step 3.1, the coefficient of $s_{\nu }$ in $s_{\lambda }{s}_{\mu }$ is the number of admissible tableaux $T$ with $\lambda +\mathrm{wt}(T)=\nu$. The Littlewood--Richardson theorem [F4] says that this same Schur coefficient is $c_{\lambda \mu }^{\nu }$, the number of LR tableaux of shape $\nu \mathrm{/}\lambda$ and content $\mu$. Thus the two counts agree.

Tensor multiplicities. Combining steps 3.1 and 4.1, the coefficient of $s_{\nu }$ in $s_{\lambda }{s}_{\mu }$ is $c_{\lambda \mu }^{\nu }$ for every partition $\nu$ with $\mathrm{\ell }(\nu )\le r$. Since $\mathrm{ch}(S_{\lambda }(V)\otimes S_{\mu }(V))=s_{\lambda }{s}_{\mu }$ and this tensor product is completely reducible with linearly independent Schur characters [F3], the multiplicity of $S_{\nu }(V)$ in $S_{\lambda }(V)\otimes S_{\mu }(V)$ is the coefficient of $s_{\nu }$, namely $c_{\lambda \mu }^{\nu }$; the terms with $\mathrm{\ell }(\nu )>r$ do not occur because $S_{\nu }(V)=0$ there.