Can we get the formula for $\prod\limits_{k=0}^n{(1+2^k)^2}$ in terms of $n$? Can we get the formula for $\prod\limits_{k=0}^n{(1+2^k)^2}$ in terms of $n$?
 A: As Cameron Williams pointed out, it suffices to consider the non-squared version. But note that
$$ \prod_{k=0}^{n} (2^{k} + 1) = \prod_{k=0}^{n} 2^{k}(1 + 2^{-k}) = \left( \prod_{k=0}^{n} 2^{k} \right)\left( \prod_{k=0}^{n} (1 + 2^{-k}) \right). $$
Taking logarithm, we find that
$$ \log \prod_{k=0}^{n} 2^{k} = \sum_{k=0}^{n} k \log 2 = \frac{n(n+1)}{2}\log 2
\quad \Longrightarrow \quad \prod_{k=0}^{n} 2^{k} = 2^{n(n+1)/2}. $$
On the other hand,
$$ \log \prod_{k=0}^{n} (1 + 2^{-k}) = \sum_{k=0}^{n} \log(1 + 2^{-k}), $$
which converges as $n \to \infty$ by limit comparison test with $\sum_{k=0}^{n} 2^{-k}$. So if we denote
$$C = \prod_{k=}^{\infty} (1 + 2^{-k}),$$
then we get
$$ \prod_{k=0}^{n} (2^{k} + 1) \sim C 2^{n(n+1)/2}. $$
A relatively fast-converging series expansion for $\log C$ can be obtained by Fubini's Theorem:
$$ \log C = \log 2 + \sum_{j=1}^{\infty} \frac{(-1)^{j-1}}{j(2^{j} - 1)} \approx 4.7684620580627434483 \cdots. $$
A: Here is all that is known about this sequence on OEIS. (Or rather, the sequence before squaring it.) I see no formulas that are not recursive or do not involve $\sum$ or $\prod$.
