The twelfth question: Find all pairs of positive integers $(k, n)$ that satisfy $k!=\prod_{\mathrm{i}=0}^{\mathrm{n}-1}\left(2^{\mathrm{n}}-2^{\mathrm{i}}\right)$.

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Here is the translation, maintaining the original text's line breaks and format.
The twelfth question:
Solution: First, prove a lemma: when the positive integer $n \geq 5$, $\left[\frac{n(n-1)}{2}+1\right]!>\prod_{i=0}^{n-1}\left(2^{n}-2^{i}\right)$.
Proof of the lemma: When $n=5$, it is verified by calculation that the conclusion holds.
Assume the conclusion holds for $n (n \geq 5)$, and consider the case for $n+1$. Note that:
$$
\begin{array}{l}
{\left[\frac{n(n+1)}{2}+1\right]!=\prod_{j=1}^{n}\left(\frac{n(n-1)}{2}+1+i\right) \cdot\left[\frac{n(n-1)}{2}+1\right]!} \\
\prod_{i=0}^{n}\left(2^{n+1}-2^{i}\right)=\left(2^{n+1}-1\right) \cdot 2^{n} \cdot \prod_{i=0}^{n-1}\left(2^{n}-2^{i}\right)
\end{array}
$$

According to the induction hypothesis, we only need to prove:
$$
\prod_{j=1}^{n}\left(\frac{n(n-1)}{2}+1+i\right) \geq\left(2^{n+1}-1\right) \cdot 2^{n}
$$

In fact,
$$
\prod_{i=1}^{n}\left(\frac{n(n-1)}{2}+1+1\right) \geqslant\left(\frac{n(n-1)}{2}+2\right)^{n} \geq\left(\frac{5(5-1)}{2}+2\right)^{n}=12^{n}>2 \times 4^{n}>\left(2^{n+1}-1\right) \cdot 2^{n}
$$

Thus (*) holds, and the lemma is proved!
Now, let's return to the original problem.
According to the conditions, we have:
$$
k!=\prod_{i=0}^{n-1}\left(2^{n}-2 i\right)=\prod_{i=0}^{n-1} 2^{i}\left(2^{n-i}-1\right)=2^{\frac{n(n-1)}{2}} \cdot \prod_{i=0}^{n-1}\left(2^{n-i}-1\right)
$$

Thus, $v_{2}(k!)=k-S_{2}(k)=\frac{n(n-1)}{2}$, and we know $k \geq \frac{n(n-1)}{2}+S_{2}(k) \geq \frac{n(n-1)}{2}+1$.
When $n \geq 5$, according to the lemma, we have:
$$
k!\geq\left[\frac{n(n-1)}{2}+1\right]!\geq \prod_{i=0}^{n-1}\left(2^{n}-2^{i}\right)
$$

This contradicts the given condition. Therefore, $n \leq 4$.
When $n=1, 2, 3, 4$, upon inspection, only $(k, n)=(1,1), (3,2)$ satisfy the conditions.