Let $k$ be a natural number.Find all the couples of natural numbers $(n,m)$ such that :
$(2^k)!=2^n*m$
1. **Taking the 2-adic valuation**:
   We start by taking the 2-adic valuation $\nu_2$ on both sides of the equation $(2^k)! = 2^n \cdot m$. This gives us:
   \[
   \nu_2((2^k)!) = \nu_2(2^n \cdot m)
   \]
   
2. **Applying Legendre's formula**:
   According to Legendre's formula, the 2-adic valuation of $(2^k)!$ is given by:
   \[
   \nu_2((2^k)!) = 2^k - S_2(2^k)
   \]
   where $S_2(2^k)$ is the sum of the digits in the binary representation of $2^k$. Since $2^k$ is a power of 2, its binary representation is a single 1 followed by $k$ zeros, so $S_2(2^k) = 1$. Therefore:
   \[
   \nu_2((2^k)!) = 2^k - 1
   \]

3. **Equating the 2-adic valuations**:
   On the right-hand side, we have:
   \[
   \nu_2(2^n \cdot m) = \nu_2(2^n) + \nu_2(m) = n + \nu_2(m)
   \]
   Since $\nu_2((2^k)!) = 2^k - 1$, we get:
   \[
   2^k - 1 = n + \nu_2(m)
   \]

4. **Solving for \( n \)**:
   Let $\nu_2(m) = t$ where $t \geq 0$. Then:
   \[
   2^k - 1 = n + t \implies n = 2^k - 1 - t
   \]

5. **Substituting \( n \) back into the original equation**:
   Substitute $n = 2^k - 1 - t$ into the original equation $(2^k)! = 2^n \cdot m$:
   \[
   (2^k)! = 2^{2^k - 1 - t} \cdot m
   \]
   Solving for $m$, we get:
   \[
   m = \frac{(2^k)!}{2^{2^k - 1 - t}}
   \]

6. **Conclusion**:
   Therefore, the pairs $(n, m)$ that satisfy the given equation are:
   \[
   (n, m) = \left(2^k - 1 - t, \frac{(2^k)!}{2^{2^k - 1 - t}}\right) \text{ for some } t \in \mathbb{Z}_{\geq 0}
   \]

The final answer is \(\boxed{(n, m) = \left(2^k - 1 - t, \frac{(2^k)!}{2^{2^k - 1 - t}}\right) \text{ for some } t \in \mathbb{Z}_{\geq 0}}\)