$$\forall{n,a>1}:\;\sum\limits\_{k=1}^{2^n-1}\frac{1}{k^a}\;\leq\left(\frac{1-2^{n(1-a)}}{1-2^{1-a}}\right)$$

For any fixed value of $a > 1$.

Induction step:

$$\sum\_{k=1}^{2^{n+1} - 1} \frac{1}{k^a} = (\sum\limits\_{k=1}^{2^n-1}\frac{1}{k^a}) + \frac{1}{(2^n)^a} + \frac{1}{(2^n + 1)^a} + \frac{1}{(2^n + 2)^a} + ... + \frac{1}{(2^{n+1} -1)^a} \leq\left(\frac{1-2^{n(1-a)}}{1-2^{1-a}}\right) + \frac{1}{(2^n)^a} + \frac{1}{(2^n + 1)^a} + \frac{1}{(2^n + 2)^a} + ... + \frac{1}{(2^{n+1} -1)^a}$$

I need help from Induction step and on. So if someone would help me, that would be greatly appreciated! People on this website keep putting this problem on hold even though I have clarified it as much as I can.

I need to prove P(n+1) is true:

$$\forall{n,a>1}:\;\sum\limits\_{k=1}^{2^{n+1}-1}\frac{1}{k^a}\;\leq\left(\frac{1-2^{(n+1)(1-a)}}{1-2^{1-a}}\right)$$