Homogenization and dehomogenization I want to assess the following claim:

If $F \in k[x_{0},...,x_{n}]$ is homogeneous and $F=x_{0}^{\alpha}G$ (where $x_{0}$ does not divide $G$) then $\beta \circ \alpha(F)=G$.

Here $\alpha$ denotes the dehomogeneization map with respect to $x_0$: $$\alpha(f) = f(1, x_1, \ldots, x_n),$$while $\beta$ is the homogeneization map for the same variable: $$\beta(f) = x_0^{d(G)} f(x_0/x_0, \ldots, x_n/x_0).$$Here $d(f)$ denotes the degree of the polynomial $f$.
Attempt: I tried with the following:
$$\beta \circ \alpha(F)(x_{0},x_{1},...,x_{n})=x_{0}^{d(G)}G(1,\frac{x_{1}}{x_{0}},\frac{x_{2}}{x_{0}},...,\frac{x_{n}}{x_{0}}).$$
So why do we have the equality $x_{0}^{d(G)}G(1,\frac{x_{1}}{x_{0}},\frac{x_{2}}{x_{0}},...,\frac{x_{n}}{x_{0}}) = G$?

Let's write $G = \sum_{|\alpha| = d(G)} a_\alpha x^\alpha$, we can do so as $G$ is homogeneous. Now we have 
\begin{align*}
  x_0^{d(G)}G\left(1, \frac{x_1}{x_0}, \ldots, \frac{x_n}{x_0}\right)
      &= x_0^{d(G)} \cdot \sum_{|\alpha| = d(G)} a_\alpha 1^{\alpha_0} \cdot \frac{x_1^{\alpha_1}}{x_0^{\alpha_1}} \cdots \frac{x_n^{\alpha_n}}{x_0^{\alpha_n}}\\\
  &= x_0^{d(G)} \cdot \sum_{|\alpha| = d(G)} a_\alpha 1^{\alpha_0} \cdot \frac{x_1^{\alpha_1} \cdots x_n^{\alpha_n}}{x_0^{\alpha_1 + \cdots + \alpha_n}}\\\
  &= x_0^{d(G)} \cdot \sum_{|\alpha| = d(G)} a_\alpha \frac{x_1^{\alpha_1} \cdots x_n^{\alpha_n}}{x_0^{d(G)-\alpha_0}}\\\
  &= \sum_{|\alpha| = d(G)} a_\alpha x_0^{\alpha_0} \cdot x_1^{\alpha_1} \cdots x_n^{\alpha_n}\\\
  &= \sum_{|\alpha| = d(G)} a_\alpha x^\alpha\\\
  &= G(x_0, \ldots, x_n).
\end{align*}
Hope that helps.
