limit of a quotient of sequences Consider the sequence $u_n=\prod_{k=0}^{n-1}(2^{2^n}-2^{2^k})$. I want to prove that $\lim_{n\to+\infty}\frac{u_{n+1}}{u^2_n}=+\infty$. I did not manage to prove it. For this, I write
$$\ln(u_{n+1})-2\ln(u_n)=2^{n+1}\ln(2)+\sum_{k=0}^n\ln\left(1-\frac1{2^{2^{n+1}-2^k}}\right)-2\sum_{k=0}^n\ln\left(1-\frac1{2^{2^n-2^k}}\right)$$
I do not know how to conclude. Thanks in advance for any help.

Using that $\;2^{2^{n+1}}-2^{2^{k+1}}=\left(2^{2^n}-2^{2^{k}}\right)\left(2^{2^n}+2^{2^{k}}\right)\,$:
$$
\require{cancel}
\begin{align}
\frac{u_{n+1}}{u_n^2} &= \frac{\prod_{k=0}^{n}(2^{2^{n+1}}-2^{2^k})}{\left(\prod_{k=0}^{n-1}(2^{2^n}-2^{2^k})\right)^2}
\\ &= \left(2^{2^{n+1}}-2^{2^0}\right) \cdot \frac{\prod_{k=1}^{n}(2^{2^{n+1}}-2^{2^k})}{\left(\prod_{k=0}^{n-1}(2^{2^n}-2^{2^k})\right)^2}
\\ &= \left(2^{2^{n+1}}-2\right) \cdot \frac{\prod_{k=0}^{n-1}(2^{2^{n+1}}-2^{2^{k+1}})}{\left(\prod_{k=0}^{n-1}(2^{2^n}-2^{2^k})\right)^2}
\\ &= 2 \left(2^{2^{n+1}-1}-1\right) \cdot \frac{\prod_{k=0}^{n-1}\cancel{(2^{2^{n}}-2^{2^{k}})}(2^{2^{n}}+2^{2^{k}})}{\prod_{k=0}^{n-1}\cancel{(2^{2^n}-2^{2^k})}(2^{2^n}-2^{2^k})}
\\ &= 2 \left(2^{2^{n+1}-1}-1\right) \prod_{k=0}^{n-1}\left(1 + \frac{2}{2^{2^n-2^k}-1}\right)
\\ &\ge 2 \left(2^{2^{n+1}-1}-1\right)
\end{align}
$$
