1. Lemma: $\nu_2\left( \binom{2n}{n}\right)$ is equal to the number of $1$'s in the binary representation of $n$.

2. Proof of Lemma: This is a direct application of Kummer's Theorem, which states that the power of a prime $p$ dividing a binomial coefficient $\binom{a}{b}$ is equal to the number of carries when $b$ and $a-b$ are added in base $p$. For $p=2$, this translates to the number of $1$'s in the binary representation of $n$.

3. Main Proof:
We need to prove that:
$$v_2\left(\prod_{n=1}^{2^m}\binom{2n}{n}\right)=m2^{m-1}+1.$$

4. Using the lemma, we have:
$$v_2\left(\prod_{n=1}^{2^m}\binom{2n}{n}\right) = \sum_{n=1}^{2^m} v_2\left(\binom{2n}{n}\right).$$

5. According to the lemma, $v_2\left(\binom{2n}{n}\right)$ is the number of $1$'s in the binary representation of $n$. Therefore, we need to sum the number of $1$'s in the binary representations of all integers from $1$ to $2^m$.

6. The number of $1$'s in the binary representation of numbers from $1$ to $2^m$ can be calculated as follows:
- For each bit position $k$ (from $0$ to $m-1$), there are $2^{m-1}$ numbers that have a $1$ in that position.
- Therefore, the total number of $1$'s in the binary representations of numbers from $1$ to $2^m$ is:
$$\sum_{k=0}^{m-1} 2^{m-1} = m \cdot 2^{m-1}.$$

7. However, we need to account for the fact that $2^m$ itself has exactly one $1$ in its binary representation, which adds an additional $1$ to the sum.

8. Therefore, the total sum is:
$$\sum_{n=1}^{2^m} v_2\left(\binom{2n}{n}\right) = m \cdot 2^{m-1} + 1.$$

9. Hence, we have:
$$v_2\left(\prod_{n=1}^{2^m}\binom{2n}{n}\right) = m \cdot 2^{m-1} + 1.$$

$\blacksquare$