Let $ p = 2n + 1, X = \prod_{k = 1}^n k, Y = \prod_{k = n + 1}^{2n}k, (p - 1)! = XY$. Then $ 2^{2n}(p - 1)! = \prod_{x - even}x\prod_{x - odd}(p + x).$ Therefore $ 4^n = 1 + p\sum_{x - odd}\frac 1x + p^2\sum_{i < j - odd}\frac {1}{ij} + O(p^3).$ $ 2\sum_{i < j - odd}\frac {1}{ij} = Z^2 - T,T = \sum_{k - odd}\frac {1}{k^2}, Z = \sum_{k - odd}\frac 1k.$ $ 2T = \sum_{k - odd}\frac {1}{k^2} + \sum_{k - even}\frac {1}{(p - k)^2} = O(p)$, when $ p > 3$. Therefore for p>3 $ 4^n = 1 + pZ + \frac {p^2}{2}Z^2 + O(p^3).$ $ 2^nY = \prod_{x - odd}(p + x),2^nX = \prod_{x - odd}(p - x)$, therefore $ C_{2n}^n = \prod_{x - odd}\frac {p + x}{p - x}.$ It gives $ {( - 1)^nC_{2n}^n = \prod_{x - odd}(1 + \frac px )(1 - \frac px )^{ - 1} = \prod_{x - odd}(1 + 2p\frac 1x + 2p^2\frac {1}{x^2}) + O(p^3).}$ Therefore $ {( - 1)^nC_{2n}^n = 1 + 2pZ + 2p^2T + 4p^2\sum_{i < j - odd}\frac {1}{ij} + O(p^3) = 1 + 2pZ + 2p^2Z^2 + O(p^3).}$ and $ {4^{2n} = (1 + pZ + \frac {p^2}{2}Z^2)^2 + O(p^3) = 1 + 2pZ + 2p^2Z^2 + O(p^3) = ( - 1)^nC_{2n}^n + O(p^3).}$