Find all pairs (k,n)(k, n)(k,n) of positive integers such that k!=(2n−1)(2n−2)(2n−4)⋯(2n−2n−1).k! = (2^n - 1)(2^n - 2)(2^n - 4) \cdots (2^n - 2^{n-1}).k!=(2n−1)(2n−2)(2n−4)⋯(2n−2n−1).