Homogeneity of degree zero and normalization One of the first assumption is that the demand function is homogeneous of degree zero. The reason and the proof is easy. 

It should also be easy why this implies we can normalize the price of one good to 1, but I cannot see exactly what's going on.
A counterexample would be of great help i.e. a function with homogeneity different from zero showing that the normalization is not possible/leads to wrong conclusions.
Demand $x(p, m)$ is the solution to the utility maximization problem:

$\max\limits_{x\in\mathbb{R}^n_+} \ \ u(x) \\ \text{s.t.} \ \  p\cdot x \leq m$

where $p\in\mathbb{R}^n_{++}$ is price vector, and $m$ is the income.

When we multiply both sides of the constraint in problem above by $\lambda &gt; 0$, and look at the revised problem we get

$\max\limits_{x\in\mathbb{R}^n_+} \ \ u(x) \\ \text{s.t.} \ \  \lambda p\cdot x \leq \lambda m$

Since this operation does not affect the constraint, the solution remains unaffected i.e. demand satisfy $x(\lambda p, \lambda m) = x(p,  m) $ which shows that demand is homogeneous of degree 0 in $(p, m)$. So, this is always true for demand function. Given that $p_1 &gt; 0$, we can take $\lambda  = \frac{1}{p_1}$, and find $x\left(\frac{p}{p_1}, \frac{m}{p_1}\right)$ to get $x(p, m)$. 

It is helpful to note that for any function $f(p)$ that is homogeneous of degree $k &gt; 0$, it is the case that $f(\lambda p) = \lambda^k f(p) \neq f(p)$ for $\lambda \neq 1$.