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For $n\ge 0$, define $B_0(X)=1$ and $B_n(X)=X(X-1)\cdots (X-n+1)/n!$. The binomial coefficient formula shows $B_n(m)\in \mathbb{Z}$ for every nonnegative integer $m$. Since these integers are dense in the p-adic integers, continuity of the polynomial implies $B_n({\mathbb{Z}}_p)\subseteq {\mathbb{Z}}_p$, so it is an integer-valued polynomial there and $\Vert B_n{\Vert }_{\infty }=1$. The Pascal's identity gives $\Delta B_{n+1}=B_n$ for the forward difference operator.

Let $f_r(m)$ be the lower-shadow size prescribed by the binomial representation in the Kruskal-Katona theorem. ThenInduction using Pascal's identity proves the inequality and drives the standard ground-set induction for the theorem.

A discrete antiderivative of $f$ is a function $F$ satisfying $\Delta F=f$. Pascal's identity gives

The Mahler coefficients tend to zero, so the displayed series defines a continuous function and has value zero at zero. The Pascal's identity and uniform convergence give $\Delta Jf=f$. Thus the forward difference operator is surjective on continuous functions on the p-adic integers. Its kernel consists of constants: period one implies agreement on the dense nonnegative integers, and continuity then implies constancy. This selects the unique discrete antiderivative vanishing at zero.

There is also a direct construction on locally constant functions. If $h$ has period $M=p^r$, put $a_j=h(j)$, $S={\sum }_{i=0}^{M-1}{a}_i$, and $P_j={\sum }_{i<j}{a}_i$. For $x=j+My$ with $0\le j<M$ and $y\in {\mathbb{Z}}_p$, setThis is a continuous function on each residue class. Increasing $j$ gives $Jh(x+1)-Jh(x)=a_j$; at $j=M-1$, the next point has residue zero and quotient $y+1$, giving the same identity. The ultrametric inequality gives $\Vert Jh{\Vert }_{\infty }\le \Vert h{\Vert }_{\infty }$, and $Jh(0)=0$. The construction is independent of the chosen period, since two normalized discrete antiderivatives agree on the nonnegative integers and then on the p-adic integers by continuity. It is linear on the locally constant functions, which form a dense subset in the supremum norm. Completeness therefore extends it to every continuous function, retaining $\Delta J=I$ and the norm bound. This proves surjectivity without first using the Mahler theorem.

For nonnegative integers $n,m$, the binomial coefficients satisfyThe identity follows by mathematical induction: adding the next term combines two adjacent binomial coefficients through Pascal's identity. An equivalent form is ${\sum }_{r=j}^N\left(\genfrac{}{}{0ex}{}{r}{j}\right)=\left(\genfrac{}{}{0ex}{}{N+1}{j+1}\right)$, a diagonal sum in Pascal's triangle. It provides closed forms for sums of binomial coefficients without expanding individual factorials.

For a nonempty $r$-uniform set family, define $x>r-1$ by $|\mathcal{F}|=\left(\genfrac{}{}{0ex}{}{x}{r}\right)$ using real binomial coefficients. Then its lower shadow has at least $\left(\genfrac{}{}{0ex}{}{x}{r-1}\right)$ members. This continuous estimate is often easier to use than the exact integer expansion in the Kruskal-Katona theorem.

A short proof uses coordinate shifts of a set family. In a fully left-shifted set family split at coordinate one into ${\mathcal{F}}_0$ and ${\mathcal{F}}_1$, deleting one in the present section. The shifting property implies $\partial {\mathcal{F}}_0\subseteq {\mathcal{F}}_1$, and the total lower shadow has size $|{\mathcal{F}}_1|+|\partial {\mathcal{F}}_1|$. Induct on uniformity and, at fixed uniformity, on set family size. If $|{\mathcal{F}}_1|<\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, the Pascal's identity and the size mathematical induction force $|\partial {\mathcal{F}}_0|>\left(\genfrac{}{}{0ex}{}{x-1}{r-1}\right)$, a contradiction. The uniformity mathematical induction applied to ${\mathcal{F}}_1$, followed by the Pascal's identity, proves the estimate. Integer $x$ is sharp for the full $r$-level on $x$ coordinates.

Row $n$ of Pascal's triangle consists of the binomial coefficients $\left(\genfrac{}{}{0ex}{}{n}{0}\right),\dots ,\left(\genfrac{}{}{0ex}{}{n}{n}\right)$, beginning with row $0$. Boundary entries are one, and Pascal's identity says each interior entry is the sum of the two entries above it. The hockey-stick identity evaluates sums along its diagonals.