Why is it that energy of electrons in higher orbits is greater than the ones in lower orbits?
What I'm having trouble understanding is why the energy level is
  higher and not lower. The electrons are further and further away from
  the nucleus, at infinity their energy would be zero.


You are correct. The innermost electrons are the most tightly bound to the nucleus and the outermost ('valence') electrons have the least energy, explaining why they can be removed (in chemical reactions e.g.)

But these energies are always negative and use the principle that infinitely far away from the nucleus the electron's energy would be $0$.

This way defined, the energy of the hydrogen electron in the ground state ($1s$) e.g. is $-13.6\ \mathrm{eV}$. The energy of the electron in excited states ($2s$, $2p$, $3s$ etc) is less negative, until for infinite distance (a free electron) it is zero ($0$).

Let's take a numerical example to demonstrate this. The ground state ($n=1$) of hydrogen's electron is $E_1=-13.6\ \mathrm{eV}$ and the first excited state ($n=2$) is $E_2=-3.4\ \mathrm{eV}$. What are the energies needed to remove the electron from these states?

Since as the energy at infinite distance is $0$, in the case $n=1$, then:

$$\Delta E= 0-(-13.6)=13.6\ \mathrm{eV}$$

And in the case $n=2$, then:

$$\Delta E= 0-(-3.4)=3.4\ \mathrm{eV}$$

So it takes less energy to remove an electron in a higher orbital than a lower one. That fits our observations.

Equally, the energy needed to excite the electron from $n=1$ to $n=2$ is:

$$\Delta E_{1\to 2}=-3.4-(-13.6)=10.2\ \mathrm{eV}$$