Accidentally, I proved the following formula for the Kronecker coefficients using some obscure method.
$$g\bigl(\ell^{mn}, m^{\ell n},n^{\ell m}\bigr)=1,\ \forall l,m,n\in\mathbb{N},$$
where $n^m$ is the rectangle partition.
By definition, the Kronecker coefficients $g\_{\mu,\nu}^\lambda$ is the structure coefficient in the tensor product
$$S\_\mu\otimes S\_\nu = \sum\_{\lambda} g\_{\mu,\nu}^\lambda S\_\lambda,$$
where $S\_\mu$ is the irreducible representation of the symmetric group defined by the partition $\mu$.
Is this a known result or maybe just some exercise?



A formula on Kronecker coefficients
I have not seen this particular statement appearing in the literature, but here is an easy way to see why it is true:


The Kronecker coefficients $g(\lambda,\mu,\nu)$ (note that such notation makes sense since these coefficients are actually invariant under any permutation of $\lambda,\mu,\nu$) can be interpreted as the coefficient of $s\_{\lambda}(x)s\_{\mu}(y)$ in $s\_{\nu}(xy)$:
$$(\*) \qquad s\_{\nu}(x.y) = \sum\_{\lambda,\mu} g(\lambda,\mu,\nu)s\_{\lambda}(x)s\_{\mu}(y).$$
Here $s$ are the Schur functions, $x=(x\_1,x\_2,\ldots)$ and $y=(y\_1,y\_2,\ldots)$ are variables and $x.y = (x\_1y\_1,x\_1y\_2,\ldots,x\_2y\_1,x\_2y\_2,\ldots)$.


In this particular case, let $\nu = m^{n\ell}$, $\lambda = (\ell m)^n$ and $\mu = (mn)^\ell$, and $x=(x\_1,\ldots,x\_n)$, $y=(y\_1,\ldots,y\_\ell)$. It is easy to see that for any $a,b$, $$s\_{a^b}(z\_1,\ldots,z\_b) =(z\_1\cdots z\_b)^a.$$
(there is only one SSYT of shape $a^b$ filled with at most $b$ distinct numbers)


Then $x.y$ consists of $n\ell$ variables, so 
$$s\_{m^{n\ell}}(x.y) = (\prod\_{i=1}^n \prod\_{j=1}^\ell (x\_i y\_j) )^m = \prod\_i x\_i^{m\ell} \cdot \prod\_j y\_j^{mn} =s\_{ (m\ell)^n} (x) s\_{(mn)^\ell}(y)$$
and comparing with (\*) we get the Kronecker coefficient in question. 


