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For a nonnegative integer $n$ and integer base $p\ge 2$, write its base-p expansion asThe base-$p$ digit sum is $d_p(n)={\sum }_r{\alpha }_r$, with $d_p(0)=0$. When $p$ is a prime number, the Legendre formula becomesTo see the connection, divide the base-p expansion by $p^j$ and discard the fractional part. Swapping the resulting finite sums and using a finite geometric series givesIn particular, $50=1\cdot 27+2\cdot 9+1\cdot 3+2=(1212{)}_3$, so $d_3(50)=6$ and $v_3(50!)=(50-6)/2=22$. The leading numeral digit contributes thirteen, the next contributes eight, the next one, and the units contribute zero. Thus the base-p digit sum compresses the same contribution ledger into information already present in positional notation.

For an integer $a$ and positive integer $b$, Euclidean division writes $a=bq+r$ with unique integer quotient $q$ and remainder $0\le r<b$. For $a\ge 0$, repeated quotient-and-remainder steps yield the digits of a base-p expansion when $b=p$. The uniqueness follows because the difference of two permitted remainders cannot be a nonzero multiple of $b$.

For a prime number $p$ and integers $0\le k\le n$, the P-adic valuation of the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ equals the number of arithmetic carries in the base-$p$ addition of $k$ and $n-k$.^[ref] Indeed, subtract the two denominator factorial valuations in the Legendre formula:Each arithmetic carry accounts for one unit of this quotient by subadditivity of the base-p digit sum. In particular, the binomial coefficient is divisible by $p$ exactly when at least one carry occurs.

For $n=50,k=17,p=3$, the factorial counts giveThe matching base-p expansions are $17=(0122{)}_3$ and $33=(1020{)}_3$. Their units add to two; the next column gives $2+2=4=1+3$, sending one carry into the nines column. That column then gives $1+0+1=2$, and no further carry occurs. The result is $(1212{)}_3=50$. The base-p digit sums check the same result: $(5+3-6)/(3-1)=1$. Consequently $\left(\genfrac{}{}{0ex}{}{50}{17}\right)$ is divisible by three but not by nine.

For a prime number $p$, the binomial coefficient modulo $p$ factors into the corresponding digit binomial coefficients in the base-p expansions of the nonnegative integers $n,k$. Pad both expansions to the same length and use $\left(\genfrac{}{}{0ex}{}{a}{b}\right)=0$ when $b>a$. The identity follows by factoring $(1+x{)}^n$ into digit powers and using the Frobenius endomorphism; digit uniqueness identifies each coefficient.

Write both nonnegative integers $n,k$ in base-p expansions, padding with zero digits to a common length $L$. By the preceding polynomial identity,Expanding the last product with the binomial theorem, each term is obtained by choosing an integer $j_i$ with $0\le j_i\le n_i<p$. Its exponent is ${\sum }_i{j}_i{p}^i$ and its coefficient is ${\prod }_i\left(\genfrac{}{}{0ex}{}{n_i}{j_i}\right)$. Uniqueness of base-p expansions says that the only way this exponent can equal $k$ is to choose $j_i=k_i$ for every $i$.

If some $k_i>n_i$, no such term exists and the coefficient is zero. Otherwise its coefficient is precisely the product below. Comparing the coefficient of $x^k$ proves Lucas theorem:with the convention $\left(\genfrac{}{}{0ex}{}{a}{b}\right)=0$ for $b>a$. This also covers $k>n$, and padding either expansion by further zero digits multiplies the product only by $\left(\genfrac{}{}{0ex}{}{0}{0}\right)=1$.

Suppose two base-p expansions represent the same nonnegative integer. Pad the shorter one by trailing zero digits so that both have the same length. Reducing modulo $p$ shows that their first digits satisfy $k_0\equiv k_0^{\prime }(\mathrm{mod}p)$. Since both digits lie between $0$ and $p-1$, they are equal as integers.

Subtract that common digit and divide by $p$. The remaining equality is an equality of two base-p expansions with one fewer digit. Repeating this argument shows that every pair of corresponding digits agrees. Equivalently, repeated Euclidean division recovers the digits as remainders. The digits are unique up to padding by zeroes. The finite nonnegative-digit representation presupposes $k\ge 0$; a negative integer has no such expansion.