[Number Theory]
Problem: Find all pairs $(m, n)$ of positive integers satisfying the equation

$$
\left(2^{n}-1\right)\left(2^{n}-2\right)\left(2^{n}-4\right) \cdots\left(2^{n}-2^{n-1}\right)=m !
\tag{1}
$$

Solution: For any prime $p$ and positive integer $N$, we will denote by $v_{p}(N)$ the exponent of the largest power of $p$ that divides $N$. The left-hand side of (1) will be denoted by $L_{n}$; that is, $L_{n}=\left(2^{n}-1\right)\left(2^{n}-2\right)\left(2^{n}-4\right) \cdots\left(2^{n}-2^{n-1}\right)$.

We will get an upper bound on $n$ from the speed at which $v_{2}\left(L_{n}\right)$ grows.

From

$$
L_{n}=\left(2^{n}-1\right)\left(2^{n}-2\right) \cdots\left(2^{n}-2^{n-1}\right)=2^{1+2+\cdots+(n-1)}\left(2^{n}-1\right)\left(2^{n-1}-1\right) \cdots\left(2^{1}-1\right)
$$

we read

$$
v_{2}\left(L_{n}\right)=1+2+\cdots+(n-1)=\frac{n(n-1)}{2}
$$

On the other hand, $v_{2}(m !)$ is expressed by the Legendre formula as

$$
v_{2}(m !)=\sum_{i=1}^{\infty}\left\lfloor\frac{m}{2^{i}}\right\rfloor
$$

As usual, by omitting the floor functions,

$$
v_{2}(m !)<\sum_{i=1}^{\infty} \frac{m}{2^{i}}=m
$$

Thus, $L_{n}=m$ ! implies the inequality

$$
\frac{n(n-1)}{2}<m
\tag{2}
$$

In order to obtain an opposite estimate, observe that

$$
L_{n}=\left(2^{n}-1\right)\left(2^{n}-2\right) \cdots\left(2^{n}-2^{n-1}\right)<\left(2^{n}\right)^{n}=2^{n^{2}}
$$

We claim that

$$
2^{n^{2}}<\left(\frac{n(n-1)}{2}\right) ! \text { for } n \geqslant 6
\tag{3}
$$

For $n=6$ the estimate $(3)$ is true because $2^{6^{2}}<6.9 \cdot 10^{10}$ and $\left(\frac{n(n-1)}{2}\right) !=15 !>1.3 \cdot 10^{12}$.

For $n \geqslant 7$ we prove (3) by the following inequalities:

$$
\begin{aligned}
\left(\frac{n(n-1)}{2}\right) ! & =15 ! \cdot 16 \cdot 17 \cdots \frac{n(n-1)}{2}>2^{36} \cdot 16^{\frac{n(n-1)}{2}-15} \\
& =2^{2 n(n-1)-24}=2^{n^{2}} \cdot 2^{n(n-2)-24}>2^{n^{2}}
\end{aligned}
$$



Putting together (2) and (3), for $n \geqslant 6$ we get a contradiction, since

$$
L_{n}<2^{n^{2}}<\left(\frac{n(n-1)}{2}\right) !<m !=L_{n}
$$

Hence $n \geqslant 6$ is not possible.

Checking manually the cases $n \leqslant 5$ we find

$$
\begin{gathered}
L_{1}=1=1 !, \quad L_{2}=6=3 !, \quad 5 !<L_{3}=168<6 ! \\
7 !<L_{4}=20160<8 ! \quad \text { and } \quad 10 !<L_{5}=9999360<11 !
\end{gathered}
$$

So, there are two solutions:

$$
(m, n) \in\{(1,1),(3,2)\}
$$
For any prime $p$ and positive integer $N$, we will denote by $v_{p}(N)$ the exponent of the largest power of $p$ that divides $N$. The left-hand side of (1) will be denoted by $L_{n}$; that is, $L_{n}=\left(2^{n}-1\right)\left(2^{n}-2\right)\left(2^{n}-4\right) \cdots\left(2^{n}-2^{n-1}\right)$.


Checking manually the cases $n \leqslant 5$ we find

$$
\begin{gathered}
L_{1}=1=1 !, \quad L_{2}=6=3 !, \quad 5 !<L_{3}=168<6 ! \\
7 !<L_{4}=20160<8 ! \quad \text { and } \quad 10 !<L_{5}=9999360<11 !
\end{gathered}
$$

We will exclude $n \geqslant 5$ by considering the exponents of 3 and 31 in (1).

For odd primes $p$ and distinct integers $a, b$, coprime to $p$, with $p \mid a-b$, the Lifting The Exponent lemma asserts that

$$
v_{p}\left(a^{k}-b^{k}\right)=v_{p}(a-b)+v_{p}(k) .
$$

Notice that 3 divides $2^{k}-1$ if only if $k$ is even; moreover, by the Lifting The Exponent lemma we have

$$
v_{3}\left(2^{2 k}-1\right)=v_{3}\left(4^{k}-1\right)=1+v_{3}(k)=v_{3}(3 k)
$$

Hence,

$$
v_{3}\left(L_{n}\right)=\sum_{2 k \leqslant n} v_{3}\left(4^{k}-1\right)=\sum_{k \leqslant\left\lfloor\frac{n}{2}\right\rfloor} v_{3}(3 k)
$$

Notice that the last expression is precisely the exponent of 3 in the prime factorisation of $\left(3\left\lfloor\frac{n}{2}\right\rfloor\right)$ !. Therefore

$$
v_{3}(m !)=v_{3}\left(L_{n}\right)=v_{3}\left(\left(3\left\lfloor\frac{n}{2}\right\rfloor\right) !\right)
$$
$$
3\left\lfloor\frac{n}{2}\right\rfloor \leqslant m \leqslant 3\left\lfloor\frac{n}{2}\right\rfloor+2 .
\tag{4}
$$

Suppose that $n \geqslant 5$. Note that every fifth factor in $L_{n}$ is divisible by $31=2^{5}-1$, and hence we have $v_{31}\left(L_{n}\right) \geqslant\left\lfloor\frac{n}{5}\right\rfloor$. Then

$$
\frac{n}{10} \leqslant\left\lfloor\frac{n}{5}\right\rfloor \leqslant v_{31}\left(L_{n}\right)=v_{31}(m !)=\sum_{k=1}^{\infty}\left\lfloor\frac{m}{31^{k}}\right\rfloor<\sum_{k=1}^{\infty} \frac{m}{31^{k}}=\frac{m}{30}
\tag{5}
$$

By combining (4) and (5),

$$
3 n<m \leqslant \frac{3 n}{2}+2
$$

so $n<\frac{4}{3}$ which is inconsistent with the inequality $n \geqslant 5$.

Final answer: $(1,1), (3,2)$