For prime \(p\) and \(k \leq n\),

Proof. By Kummer’s theorem, \(v_p\binom {n}{k}\) counts carries among at most \(\texttt{trunc\_ log}(p, n)\) positions.

For prime \(p\) with \(n {\lt} p\) and \(k \leq n\),

Proof. When \(n {\lt} p\), \(\texttt{trunc\_ log}(p, n) = 0\), so there are no carry positions.

For prime \(p\) with \(p^2 {\gt} n\) and \(k \leq n\),

Proof. Since \(p^2 {\gt} n\), \(\texttt{trunc\_ log}(p, n) \leq 1\), so there is at most one carry position.

For prime \(p\), \(k \leq n\), and \(0 {\lt} k\),

Proof.

For prime \(p\), \(0 {\lt} k \leq p^m\),

Proof. Combine Kummer’s theorem with Lemma 1.4.

For \(n {\gt} 0\),

Proof. By Legendre’s formula, \(v_2(n!) = \sum _{k \geq 1} \lfloor n/2^k \rfloor {\lt} n \cdot \sum _{k \geq 1} 2^{-k} = n\).