Consider $\mathbb{R}^n$ with the standard Euclidean inner-product. I'm trying to give a proof that
$$ \left< v, v \right> = \left|\left| v \right|\right|^2 $$
but can only seem to do it for $\mathbb{R}^2$ where you can use the geometric definition of the dot product, namely
$$ u \cdot v = \left|\left|u\right|\right|\left|\left|v\right|\right|\cos(\theta). $$
Any idea how to generalize the proof for $n > 2$?
On
The only sensible interpretation would be in the sense of the polarization formulas. That is, the formula for the norm is given and one exploits it to find related quantities.
The most basic geometric construction that is related to the norm is the projection of a point onto a line. Given points $A$ and $B$ and a direction $v$, find the point closest to $B$ on the line $A+tv$. Using the definition of the euclidean norm, this leads to a quadratic equation in $t$ where the square can be again completed,
\begin{align} \|A+tv-B\|^2 & =\|tv-AB\|^2=\sum_i\left[t^2v_i^2-2tv_i(AB)_i+(AB)_i^2\right]\\ & = t^2\|v\|^2-2t\sum_iv_i(AB)_i+\|AB\|^2\\ & = \left(t\|v\|-\frac{\sum_iv_i(AB)_i}{\|v\|}\right)^2+\|AB\|^2-\left(\frac{\sum_iv_i(AB)_i}{\|v\|}\right)^2 \end{align}
So the remaining sum inside the last line seems to be an important quantity, give it the symbol $v\cdot AB$ or $\langle v,\,AB\rangle$ and explore further uses. Turns out it is rather ubiquitous.
If your definition is the inner product space definition, that is $||v||$ is defined as $\sqrt{\langle v,v\rangle}$, then the problem is instantly trivial.
If your definition is the coordinate definition, that is $||v||$ is defined as $\sqrt{\sum\limits_{i=1}^{n}v_{i}^{2}}$, and $\langle u,v\rangle$ is defined as $\sqrt{\sum\limits_{i=1}^{n}u_{i}v_{i}}$, then the problem is instantly trivial.
If your definition is the geometric definition, that is you first impose an Euclidean plane that contains all the involved vector and then use the angle, then your proof reduced to the $\mathbb{R}^{2}$ case.