Why is the principal energy of an electron lower for excited electrons in a higher energy state?
Notice that when $n=1$, we have,
$$
E=-E_0=-13.6~\mathrm{eV}
$$
which is the negative of the energy required to remove an electron from the ground state of a hydrogen atom.

If we increase $n$ to say $n=2$, then we have,

$$
E=-E_0/4=-3.4~\mathrm{eV}
$$
which is a larger number than for $n=1$. Don't let the minus sign confuse you.

This is a very common source of confusion when these equations are seen for the first time. The confusion often stems from the fact that we are free to choose the zero of energy wherever we would like. So, in this case, zero energy corresponds to the case where the electron and nucleus are infinitely separated which is the $n\rightarrow\infty$ limit. So, more negative numbers correspond to lower energies and more tightly bound electrons.