Prove using induction principles $$\forall{n,a>1}:\;\sum\limits_{k=1}^{2^n-1}\frac{1}{k^a}\;\leq\left(\frac{1-2^{n(1-a)}}{1-2^{1-a}}\right)$$
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:
$$\forall{n,a>1}:\;\sum\limits_{k=1}^{2^{n+1}-1}\frac{1}{k^a}\;\leq\left(\frac{1-2^{(n+1)(1-a)}}{1-2^{1-a}}\right)$$

Hint. $$\sum\limits_{k=2^n}^{2^{n+1}-1} \frac{1}{k^a} \leq \sum\limits_{k=2^n}^{2^{n+1}-1}\frac{1}{2^{na}}=\frac{1}{2^{n(a-1)}}$$
and
$$
\frac{1 - 2^{n(1-a)}}{1-2^{1-a}} + \frac{1}{2^{n(a-1)}}  = \cdots
$$
By the way, a circuitous route might by using the geometric series formula (you certainly don't need to do this, but it's interesting), you know that 
$$
\sum\limits_{k=0}^{n-1} \frac{1}{2^{k(a-1)}} = \frac{1-2^{n(1-a)}}{1-2^{1-a}}
$$
and so 
$$
\frac{1 - 2^{n(1-a)}}{1-2^{1-a}} + \frac{1}{2^{n(a-1)}} = \sum\limits_{k=0}^{n-1} \frac{1}{2^{k(a-1)}} + \frac{1}{2^{n(a-1)}} = \sum\limits_{k=0}^n \frac{1}{2^{k(a-1)}} = \cdots
$$
