Determine convergence of the sequence $x_0=1 , x_{n+1}=x_n (1+ 2^{-(n+1)})$ I want to check if the following sequence converges:

$$x_0=1 , x_{n+1}=x_n \left(1+ \frac{1}{2^{n+1}}\right)$$

I proved the sequence is increasing :
$\cfrac{x_{n+1}}{x_n}=1+ \cfrac{1}{2^{n+1}} \gt 1$
Now I should prove it is bounded above. let's write some terms of the equation:
\begin{align}
x_0&=1 \\[2ex]
x_1&=1\cdot\left(1+\dfrac{1}{2^1}\right)\\[2ex]
x_2&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\\[2ex]
x_3&=\left(1+\dfrac{1}{2^1}\right)\cdot\left(1+ \dfrac{1}{2^2}\right)\cdot\left(1+ \dfrac{1}{2^3}\right)\\[2ex]
\end{align}
So, we can write:
$$x_{n+1}=\left(1+\frac{1}{2^1}\right)\cdot\left(1+ \frac{1}{2^2}\right)\cdots\left(1+\frac{1}{2^{n}}\right)\cdot\left(1+\frac{1}{2^{n+1}}\right)$$
Here, I'm not sure how to prove it is bounded above.

The Limit Exists
Cross-multiplication and comparison shows that
$$
\frac{1+\frac1{2^{k-1}}}{1+\frac1{2^k}}\le1+\frac1{2^k}\le\frac{1-\frac1{2^k}}{1-\frac1{2^{k-1}}}\tag1
$$
Then telescoping products lead to
$$
\underbrace{\prod_{k=1}^\infty\frac{1+\frac1{2^{k-1}}}{1+\frac1{2^k}}}_2
\le\prod_{k=1}^\infty\left(1+\frac1{2^k}\right)
\le\underbrace{\overset{\substack{k=1\\\downarrow\\[6pt]\,}}{\frac32}\overbrace{\prod_{k=2}^\infty\frac{1-\frac1{2^k}}{1-\frac1{2^{k-1}}}}^2}_3\tag2
$$
Therefore, $x_n$ is an increasing sequence that is bounded above by $3$, so
$$
\lim_{n\to\infty}x_n=\prod_{k=1}^\infty\left(1+\frac1{2^k}\right)\tag3
$$
exists.

Bounding The Limit
Using $(1)$ and we get
$$
\left(1+\frac1{2^n}\right)\prod_{k=1}^n\left(1+\frac1{2^k}\right)
\le\prod_{k=1}^\infty\left(1+\frac1{2^k}\right)
\le\frac1{1-\frac1{2^n}}\prod_{k=1}^n\left(1+\frac1{2^k}\right)\tag4
$$
The greater the $n$ used, the tighter the bounds in $(4)$.

The Limit
Using $(4)$ with $n=30$, we get that
$$
2.38423102903137172\color{#C00}{35}\le\prod_{k=1}^\infty\left(1+\frac1{2^k}\right)\le2.38423102903137172\color{#090}{55}\tag5
$$
