Proof by induction with summation

I have this math problem that I am stuck on.

Prove by induction on $n$ that
$$\sum_{j=1}^{2^n}{\frac{1}{j}>\frac{n}{2}}$$

What I have so far is check for $n=0$ (base case) $\sum_{j=1}^{1}{\frac{1}{j}} = 1 > 0$.

Then I assume $\sum_{j=1}^{2^n}{\frac{1}{j}>\frac{n}{2}}$ is true for all $n\ge 0$.

I then show it is true for $n+1$

$$\sum_{j=1}^{2^{n+1}}{\frac{1}{j}>\frac{n+1}{2}}$$
$$=\sum_{j=1}^{2^{n}}{\frac{1}{j} + \sum_{j=2^n}^{2^{n+1}}{\frac{1}{j}} >\frac{n+1}{2}}$$

$$=\frac{n}{2} + \sum_{j=2^n}^{2^{n+1}}{\frac{1}{j}} >\frac{n+1}{2}$$

$$= \sum_{j=2^n}^{2^{n+1}}{\frac{1}{j}} > \frac{1}{2}$$

I'm stuck here. thanks.