Why is the principal energy of an electron lower for excited electrons in a higher energy state? Several places state the 'principal energy of an electron' can be calculated as such:

$$E = \frac{2π^2mZ^2e^4}{n^2h^2}$$

Another equation I found was:

$$E = -\frac{E_0}{n^2},$$

where $$E_0 = \pu{13.6 eV}~(\pu{1 eV} = \pu{1.602e-19 J})$$

As seen in these equations, the greater the principal number ($n$) of the electron, the lower the principal energy $E$ of this electron.

However, the principal number $n$ is associated with higher energy states. The farther from the electron, the higher the energy state of this electron.

I seem to be fumbling the concept of 'principal energy of an electron.' What is the difference between the 'energy state' associated with the principal number, and the 'principal energy' of an electron? What exactly do the 'principal energy' equations mean? I read somewhere that it would be the energy it would take to "unbind" or ionize the electron, which would make sense, but I have not seen 'principal energy' explained as the ionization energy of an electron anywhere else.
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.