Důkaz.

Nechť ${ord}_{2} (n)$ je exponent u nejvyšší mocniny dvojky, která dělí $n$ . Využijeme známého vzorce

{ord}_{2} (n!) = \sum_{k = 1}^{\infty} ⌊ \frac{n}{2^{k}} ⌋

Zajímá nás

\begin{aligned} {ord}_{2} (C_{n}) & = {ord}_{2} (\frac{(\binom{2 n - 2}{n - 1})}{n}) \\ = {ord}_{2} (\frac{(2 n - 2)!}{n (n - 1)!^{2}}) \\ = {ord}_{2} ((2 n - 2)!) - 2 {ord}_{2} ((n - 1)!) - {ord}_{2} (n) \\ = \sum_{k = 1}^{\infty} ⌊ \frac{2 n - 2}{2^{k}} ⌋ - 2 \sum_{k = 1}^{\infty} ⌊ \frac{n - 1}{2^{k}} ⌋ - {ord}_{2} (n) \\ = \sum_{k = 1}^{\infty} (⌊ \frac{n - 1}{2^{k - 1}} ⌋ - 2 ⌊ \frac{n - 1}{2^{k}} ⌋) - {ord}_{2} (n) \\ = \sum_{k = 1}^{\infty} (\frac{n - 1}{2^{k - 1}} - {\frac{n - 1}{2^{k - 1}}} - 2 \frac{n - 1}{2^{k}} + 2 {\frac{n - 1}{2^{k}}}) - {ord}_{2} (n) \\ = \sum_{k = 1}^{\infty} (\frac{n - 1}{2^{k - 1}} - {\frac{n - 1}{2^{k - 1}}} - 2 \frac{n - 1}{2^{k}} + 2 {\frac{n - 1}{2^{k}}}) - {ord}_{2} (n) \end{aligned}