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Over the complex numbers, if $N^m=0$, thenHere the generalized binomial coefficient is $\left(\genfrac{}{}{0ex}{}{\alpha }{j}\right)=\alpha (\alpha -1)\cdots (\alpha -j+1)/j!$, with $\left(\genfrac{}{}{0ex}{}{\alpha }{0}\right)=1$. This is an exact finite identity, even when an infinite binomial series would not converge.

One can construct the required polynomial $q(z)=1+{\sum }_{j=1}^{m-1}{c}_j{z}^j$ without any infinite series. At polynomial degree $j$, the coefficient of $q(z{)}^3$ is $3c_j$ plus terms involving only earlier coefficients. Recursively choose it to agree with $1+z$ through polynomial degree $m-1$; the first coefficients are $1/3,-1/9,5/81$. The stated generalized binomial coefficients solve this recursion: near $z=0$, the binomial series for $(1+z{)}^{1/3}$ cubes to $1+z$, and truncating it changes no coefficient below polynomial degree $m$. Thus $q(z{)}^3-(1+z)$ is divisible by $z^m$ as a polynomial identity. Substituting the nilpotent matrix $N$ kills the remaining terms, all divisible by $N^m$. For a Jordan block $\lambda I+N$ with $\lambda \ne 0$, replace $N$ by $N/\lambda$ and multiply $Q$ by a scalar cube root of $\lambda$.

The uniform layer consists of the characteristic vectors of sets of size $r$ in $[n]$. It has size $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$, a binomial coefficient. Restricting polynomials to a uniform layer introduces identities absent on the entire Boolean lattice, since the coordinate sum is fixed. In particular, homogenisation on a uniform layer spans all restricted multilinear polynomials of degree at most $s\le r$ using just the degree-$s$ monomials.

For fixed nonnegative integer $n$, the row $\left(\genfrac{}{}{0ex}{}{n}{0}\right),\dots ,\left(\genfrac{}{}{0ex}{}{n}{n}\right)$ increases to its middle and then decreases. Indeed, for $0\le k<n$, the ratio of consecutive binomial coefficients is $(n-k)/(k+1)$, which is greater than, equal to, or less than $1$ according as $k$ is less than, equal to, or greater than $(n-1)/2$. Thus even $n$ has one maximum at $k=n/2$, while odd $n$ has two equal maxima at $k=(n-1)/2,(n+1)/2$.

Choose a $k$-element subset of a disjoint union of sets of sizes $m$ and $n$. Partition the choices by the number $i$ of selected elements from the first set. The resulting count is the displayed convolution of binomial coefficients, where coefficients outside their admissible ranges are zero. This also proves the formula when $k>m+n$.