So I'm kinda confused right now and I'm probably just confusing myself for no reason.
So let say we have a graph went just one vertex.
The complete graph of that one vertex will be a loop to itself.
It will have a Euler Circuit because it has a degree of two and starts and ends at the same point.
Am I right?
Also, I think it will have a Hamiltonian Circuit, right?
On
You are right. As there is only one vertex in this graph, and depending on what the graph looks like (a single vertex without an edge or a single vertex with a loop), we find that every top has even degree. It is also trivial to notice that this is a connected graph, so we deduce, by a theorem proven by Euler, that this graph contains an eulerian cyclus. Also, draw both cases and apply your definition of Eulerian cyclus to it! Convince yourself the definition applies here.
Complete graph $K_1$ has $\frac{0\cdot(-1)}{2} = 0$ edges and is Hamiltonian by convention. Also it is connected and all vertex degrees are even (I hope there is no surprise that 0 is even), therefore it is Eulerian.
If you want to consider pseudograph on 1 vertex then it is Hamiltonian and Eulerian, too, since addition/removal a loop doesn't change this properties.