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For nonnegative integers $r,n$, the alternating binomial moment satisfies $S_r(n)=0$ when $r<n$ and $S_n(n)=(-1{)}^{n}n!$. Expand the polynomial $X^r$ in falling factorials and differentiate the binomial theorem to prove these identities. Equivalently, alternating binomial coefficients implement a finite difference that annihilates polynomials of degree less than $n$.

These $n+1$ nonnegative polynomials form a basis of polynomials of degree at most $n$, and sum to one on $[0,1]$. For $0\le j\le n$,where ratios for $j=0$ equal one and terms with $k<j$ vanish. The identity follows from $\left(\genfrac{}{}{0ex}{}{n}{k}\right)(k{)}_j/(n{)}_j=\left(\genfrac{}{}{0ex}{}{n-j}{k-j}\right)$ and the binomial theorem. It gives explicit basis coefficients using the falling factorial.

For $m\ge 1$ and $n\ge 2$, the Bernstein polynomial of the monomial $x^m$ satisfiesIt follows by applying $k\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ and expanding $k=(k-1)+1$ with the binomial theorem. The coefficients sum to one; only the first tends to one as $n\to \infty$. Induction on $m$ proves uniform convergence of these monomial approximants, starting from $B_n(1)=1$.

The even part of a polynomial $P$ is $(P(z)+P(-z))/2$. It retains exactly the terms whose exponents are even. Applied to the binomial theorem, it extracts sums of even-indexed binomial coefficients.

Every odd composite integer has an Euler witness. For a nonsquarefree integer, a base congruent to $1+p$ modulo a repeated prime factor $p^2$ fails the Euler congruence by the binomial theorem. For a squarefree integer, use the Chinese remainder theorem to choose a base that is a quadratic nonresidue modulo one prime factor and $1$ modulo another. Its Jacobi symbol is $-1$, while its power is $1$ modulo the second factor.

If $D$ is a derivation of a Lie algebra, then its generalized eigenspaces satisfyApply a sufficiently large power of $D-(\lambda +\mu )I$ and use the binomial theorem together with the derivation rule.