Infinite product

Prove that for any positive number $t$ the following holds -
$$\prod_{n=1}^{\infty}\left(\frac{1}{2}+\frac{1}{2}e^{2^{-n}t}\right)=\frac{e^t-1}{t}$$
 (solved, solution to be posted)
Interesting corollary appears when we change $t\to\log t$. We have $$\log t=(t-1)\left(\prod_{n=1}^{\infty}\left(\frac{1}{2}+\frac{1}{2}t^{2^{-n}}\right)\right)^{-1}$$