From Step 1 to Step 2 the author used the inequality as shown in the picture above. The doubt that I have is why did he take $2^{n}$ instead of $2^{n-1}$ in the last term (circled one) ( because as per the formula n should be reduced by n-1 , you can also see the same for preceding terms)
Edit : Here's what I did next which makes sense but I'm still stuck with the circled logic
Edit 2 : If I follow as per the logical path here's what I get , can anyone confirm if my attempt is correct or wrong?
As you have alredy guessed this does indeed look like a small error, the inequality should be:
$\le \frac{1}{2^m} + ... + \frac{1}{2^{n-1}} = \frac{1}{2^m} (1 + \frac{1}{2} + ... + \frac{1}{2^{n-m-1}}) \le \frac{1}{2^{m}} \cdot 2 = \frac{1}{2^{m-1}}$
Note that you do not necessarily need to explicitly deal with the term:
$1 + \frac{1}{2} + ... + \frac{1}{2^{n-m-1}}$
That's because for all positive integers d we have:
$1 + \frac{1}{2} + ... + \frac{1}{2^d} \le 2$