Problem 4

Find all pairs

(k,n)

of positive integers such that

k!=(2^n-1)(2^n-2)(2^n-4)\cdots(2^n-2^{n-1}).

Step 4 of 6: Exclude n at least 6 by growth

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L_n<2^{n^2},\qquad m=\frac{n(n-1)}2,\qquad 2^{n^2}<m!\ (n\ge6)\Longrightarrow\ k!>m!>2^{n^2}>L_n

Detailed analysis

Every factor of $L_n$ is less than $2^n$ , so $L_n<2^{n^2}$ . Put $m=n(n-1)/2$ . For $n=6$ , $2^{n^2}=2^{36}<15!=m!$ . If $2^{n^2}<m!$ holds for some $n\ge6$ , then for $n+1$ the factorial gains $n$ factors, each at least $m+1\ge16$ , while the power gains only a factor $2^{2n+1}$ ; thus the inequality continues by induction. Hence $2^{n^2}<m!$ for every $n\ge6$ . But Step 3 gives $k>m$ , so $k!>m!$ , contradicting $k!=L_n<2^{n^2}<m!$ .