A very (hopefully) simple question:
So basically Hilbert Spaces are just Euclidean N-dimensional spaces with complex numbers instead of real numbers. Is that it, a fancy name just for this? (And of course the some redefinition of the mathematical properties of things related to this change.)
The notion Hilbert space falls into the realm of functional analysis. While some finite dimensional (real or complex) vector spaces are (technically) also Hilbert spaces it is mainly a term applied in the context of infinite dimensional vector spaces. Aside, some people reserve the term Hilbert space for (certain) real vector spaces and call the complex analog a unitary space. (But since you mention complex, let us ignore this.)
A Hilbert space is a $K$-vectorspace $V$ (where $K$ denotes the field of real or complex numbers) with an inner product $\langle \cdot , \cdot \rangle: V \times V \to V$ such that
for $v,v,' w \in V$ and $\lambda \in K$, and setting $||v|| = \sqrt{\langle v, v \rangle}$ for $v \in V$ one has that $(V, ||\cdot ||)$ is a complete normed vector space. (Note it is always a normed space, the complete is the additional condition; without that it is called a pre-Hilbert space).
Now, the conditions in the list are what you likely known (perhaps without the complex 'decorations') the conditions for a Euclidean space where there is the additional condition that the dimension is finite. (Also note that sometimes the third condition in the list is $\langle \lambda v , w \rangle = \overline{ \lambda} \langle v , w \rangle$, but this changes nothing it is just important it is clear with which defintion you work.)
A finite dimensional (real or complex) normed spaces is always complete, so you do not need the extra condition that it is complete.
A typical example for a (real) Hilbert space is the set of all infinite sequences $(a_i)_{i \in \mathbb{N}}$ such that the series $\sum_{i \in \mathbb{N}} a_i^2$ converges, where the inner product of two elements $(a_i)_{i \in \mathbb{N}}$ and $(b_i)_{i \in \mathbb{N}}$ is given as $\sum_{i \in \mathbb{N}} a_ib_i$, somehow directly generalizing the defintion from Euclidean spaces. (Of course if the number of summands is finite there is no convergence condition.)
This space is typically denoted $\ell_2 (\mathbb{N})$.
A Hilbert space does not have to be finite dimensional. Also, you could use $\mathbb R$ or $\mathbb C$ as the base field.
The definition I know is that a Hilbert space is a complete inner product space.
No. A Hilbert Space is a little more abstract than that. Although any N-dimensional Euclidean space will be a Hilbert Space, not every Hilbert space is necessary Euclidean.
To be a Hilbert space, the space merely needs an inner product defined, and that inner product needs to meet certain criteria, namely:
In a Euclidean (thus Hilbert) space, the inner product is the "dot product".
You can make some more detailed claims if the elements in the space are real numbers, but you can find that on the wiki article ;)
Consider the vector space $P_n(\mathbb{C})$ of complex polynomials in 1 variable of degree $\leq n$ with the inner product given by $\langle p, q\rangle=\int_a^bp(x)\overline{q(x)}dx$. This will be a $n+1$ dimensional Hilbert space. Since dimension is a complete invariant of a Hilbert space it will be $\textit{isomorphic}$ to $\mathbb{C}^{n+1}$. But that doesn't really mean that it $\textit{is}$ $\mathbb{C}^{n+1}$.
Note: you can do everything above with the real instead of complex.
Now your intuition is correct, Hilbert spaces are the most appropriate generalization of euclidean spaces $\textit{GEOMETRICALLY}$. This means for example that balls are round. However, the key with your confusion I think is that finite dimensional Hilbert spaces are just isomorphic to Euclidean spaces so it doesn't really seem like anything new. It is when you go to infinite dimensions that you get new examples.
