Prove that $\left(1+2^{-1+b^{\left(\frac{1}{b-1}\right)}}\right)^b < 1+2^{-1+b^{\left(\frac{b}{b-1}\right)}}$ for all $b>2$. While solving a bigger problem, I've reduced it to an inequality $$\left(1+2^{-1+b^{\left(\frac{1}{b-1}\right)}}\right)^b < 1+2^{-1+b^{\left(\frac{\color{red}b}{b-1}\right)}}$$for $b>2$, which looks plausible when looking at the plots. I've tried using Jensen inequality here with $x\mapsto x^{b-1}$, but not much luck.
I've also checked that the inequality works empirically with Wolfram Alpha.
Yes, this is written correctly, opposite to what some of the people in the comments below are trying to claim. I'm surprised this even needs claryfying.

This is a partial answer, at least. I'm planning to keep thinking about this tomorrow, but maybe someone else can step in and finish what I have. The argument is almost complete, but I have other things to work on right now.

First we log both sides to get the equivalent inequality
$$
b \log \left ( 1+2^{b^{\frac{1}{b-1}}-1} \right ) 
\overset{?}{<} 
\log \left ( 1+2^{b^{\frac{b}{b-1}}-1} \right )
$$
Then we factor out the dominant term of each log, and separate to get
$$
b 
\left [ 
  \log \left ( 2^{b^{\frac{1}{b-1}}-1} \right ) 
  + 
  \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right ) 
\right ]
\overset{?}{<}
\log \left ( 2^{b^{\frac{b}{b-1}}-1} \right ) 
+
\log \left ( 1 + 2^{1 - b^{\frac{b}{b-1}}} \right )
$$
Then we apply some log rules and rearrange
$$
b^{\frac{b}{b-1}} \log(2) 
- 
b \log(2) 
+ 
b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right )
\overset{?}{<}
b^{\frac{b}{b-1}} \log(2)
-
\log(2)
+ 
\log \left ( 1 + 2^{1 - b^{\frac{b}{b-1}}} \right )
$$
We can cancel the first term of each side, and swap the second terms to make them positive
$$
\log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right )
\overset{?}{<}
b \log(2) + \log \left ( 1 + 2^{1 - b^{\frac{b}{b-1}}} \right )
$$

Now, looking at the left hand side, notice $b^{\frac{1}{b-1}} \to 1$ from above. So we get the following honest upper bound on the left hand side
$$ 
\log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right )
<
\log(2) + b \log(2)
$$
In fact, this inequality is not very tight -- It turns out for $b > 4$, we have
$$\log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right ) < b \log(2)$$
but my only proof (so far) is desmos:


Looking at the right hand side, notice $b^{\frac{b}{b-1}} \to \infty$, so
$2^{1-b^{\frac{b}{b-1}}} \to 2^{1-\infty} \to 0$, so it's a good thing that
$b \log(2)$ eventually dominates the left hand side!

In summary, by actually proving the tighter inequality $\log(2) + b \log \left ( 1 + 2^{1 - b^{\frac{1}{b-1}}} \right ) < b \log(2)$, which I think is a reasonable goal, we can show that your desired inequality holds for $b > 4$ (really $3.384$).
But we're currently using $0$ as our lower bound for
$\log \left ( 1 + 2^{1-b^{\frac{b}{b-1}}} \right )$. One can see from the graph that the desired inequality holds from $2$ to $3.384$, so if that pleases you then we're done. If not, then slightly more work is needed, but I haven't spent much time thinking about this case yet.

I hope this helps ^_^
