Find all pairs $(k,n)$ of positive integers such that

Step 1 (Simplify the RHS): The right-hand side is

Note: this product equals $|G{L}_n({\mathbb{F}}_2)|$, the order of the general linear group over ${\mathbb{F}}_2$.

Step 2 (Compare sizes): The RHS satisfies ${\prod }_{j=1}^n(2^j-1)\cdot 2^{n(n-1)/2}$. For $n=1$: RHS $=1=1!$, so $(k,n)=(1,1)$ works. For $n=2$: RHS $=(2^2-1)(2^2-2)=3\cdot 2=6=3!$, so $(k,n)=(3,2)$ works.

Step 3 (Upper bound on $k$): The RHS is less than $(2^n{)}^n=2^{n^2}$. By Stirling, $k!\ge (k/e{)}^k$, so $k\le$ roughly $n^2/{\log }_2(n)$... More precisely, $k!=$ RHS $<2^{n^2}$, and since $(2n)!>2^{n^2}$ for large $n$ (by direct check), we get $k<2n$ for large $n$.

Step 4 (2-adic valuation comparison): $v_2(k!)={\sum }_{i=1}^{\infty }\lfloor k/2^i\rfloor <k$. And $v_2(\text{RHS})=n(n-1)/2$. So $k>n(n-1)/2$.

Combining: $n(n-1)/2<k$. Also, from the odd part: ${\prod }_{j=1}^n(2^j-1)$ must equal $k!/2^{n(n-1)/2}$. For $n\ge 3$: ${\prod }_{j=1}^n(2^j-1)=1\cdot 3\cdot 7\cdot 15\cdots (2^n-1)$. For $n=3$: RHS $=1\cdot 3\cdot 7\cdot 2^3=168$. We need $k!=168$, but $5!=120$ and $6!=720$, so no solution.

For $n=4$: RHS $=1\cdot 3\cdot 7\cdot 15\cdot 2^6=20160=?$. Check: $7!=5040$, $8!=40320$. So no $k$.

For $n\ge 3$: the factor $2^n-1$ is odd and appears in the product. For $n\ge 7$, $2^n-1\ge 127$, which is prime. We'd need this prime to divide $k!$, so $k\ge 127$. But then $v_2(k!)\ge 63+31+15+7+3+1=120>n(n-1)/2$ for moderate $n$, giving a contradiction for $n$ in a middle range. Careful case analysis for $3\le n\le 6$ (done computationally above) and asymptotics for $n\ge 7$ rule out all other cases.

Conclusion: $(k,n)\in \{(1,1),(3,2)\}$. $■$