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Ifthen the base-$p$ digit sum is $d_p(n)={\sum }_r{\alpha }_r$. The Legendre formula can equivalently be written

The central binomial coefficient and Legendre formula implyfor every sufficiently large real number $x$.

For $a=\pi (\sqrt{x})$, let $p_1,\dots ,p_a$ be the primes at most $\sqrt{x}$. The inclusion-exclusion principle counts unsieved positive integers by $\varphi (x,a)={\sum }_S(-1{)}^{|S|}\lfloor x/{\prod }_{i\in S}{p}_i\rfloor$. These are $1$ and the remaining primes, since every composite number has a prime factor at most its square root. Hence $\pi (x)=\varphi (x,a)+a-1$ for $x\ge 2$. This is a prime-counting identity, distinct from the Legendre formula for factorial valuations.

If $n=|\lambda |$ and $d_p(n)$ is its base-$p$ digit sum, thenThis combines the Hook-length formula, Legendre formula, the abacus divisible-hook correspondence, and the recurrence $q_r=c_r+p{q}_{r+1}$.

Bezout identity gives integers $b,c$ with $ab+nc=1$ whenever $\gcd (a,n)=1$, so $ab\equiv 1(\mathrm{mod}n)$. If $b,b^{\prime }$ are inverses, multiplying $a(b-b^{\prime })\equiv 0$ by an inverse gives $b\equiv b^{\prime }(\mathrm{mod}n)$. Multiplying Bezout identities for $a$ and $b$ shows that $\gcd (ab,n)=1$.

Wilson theorem states that $(p-1)!\equiv -1(\mathrm{mod}p)$ for prime $p$. Pair the nonzero residues with their unique multiplicative inverses. The only self-inverse residues are $1,-1$, whose product is $-1$, proving the theorem.

Counting one factor of $p$ for every multiple of $p$, another for every multiple of $p^2$, and so on proves Legendre formulaNow $22!\equiv 20!\cdot 21\cdot 22\equiv 2(20!)\equiv -1(\mathrm{mod}23)$, so $20!\equiv 11(\mathrm{mod}23)$. Also $v_5(1000!)=200+40+8+1=249$, while $v_2(1000!)>249$, so $1000!\equiv 0(\mathrm{mod}10^{249})$.

Finally,For $1\le j<p^m$, $v_p(p^m-j)=v_p(j)$, so the second factor is a $p$-adic unit. If $v_p(k)=\ell$, then

For the coefficient $a_n=1/n!$, the p-adic absolute value gives $|a_n{|}_p=p^{v_p(n!)}$. By Legendre formula,where $s_p(n)$ is the sum of the base-$p$ digits of $n$. Thus $v_p(n!)/n\to 1/(p-1)$. The Cauchy-Hadamard theorem now yieldsand therefore the radius of convergence of the p-adic exponential is

PutThe defining relation between the quotient tower of a partition and the core tower of a partition isSumming the resulting telescoping identities gives

By the Hook-length formula,The abacus divisible-hook correspondence says that the number of hooks divisible by $p^r$ is $q_r$, soIf $d_p(n)={\sum }_r{\alpha }_r$, the digit-sum form of the Legendre formula isCombining the three displayed identities proves the P-adic valuation of a symmetric-group character degree from the core tower formula

Legendre formula giveswhose summands are zero or one. If the sum is at least $k$, some nonzero summand has index $j\ge k$, whence $p^k\le p^j\le 2n$.

Grouping the von Mangoldt function by prime givesEvery prime-power contribution to $\log \left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ occurs in $\psi (2n)$ by the first part. Finally $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)={\prod }_{j=1}^n(n+j)/j\ge 2^n$, so $\psi (2n)\ge n\log 2$.

The binomial theorem givesThe central binomial coefficient is the largest of these $2n+1$ nonnegative binomial coefficients, so

For the upper bound, fix a prime number $p\le 2n$. By Legendre formula,Each summand is either zero or one. Hence the P-adic valuation satisfiesIf $p\le \sqrt{2n}$, this impliesThere are at most $\sqrt{2n}$ such prime numbers. If instead $p>\sqrt{2n}$, then $p^2>2n$, so the P-adic valuation is at most one. Multiplying the prime-power contributions therefore yields

On ${\mathbb{C}}_p$, the power series $\exp (x)={\sum }_{n\ge 0}{x}^n/n!$ has radius of convergenceIndeed, Legendre formula gives $v_p(n!)/n\to 1/(p-1)$, and the Cauchy-Hadamard theorem gives the stated radius.