Baby rudin theorem 10.7

We need this definition for the proof of the theorem:

Here is the theorem: Suppose $F$ is a $\mathscr C'$ - mapping ( that means continuously differentiability) of an open set E $\subset R^n$ into $R^n$, $0 \in E $, $F(0) = 0$, and $F'(0)$ is invertible.

Then there is a neighborhood of $0$ in $R^n$ in which a representation: $$\mathbf{F}(\mathbf{x})=B_1\cdots B_{n-1}\mathbf{G}_n\circ \cdots \mathbf{G}_1(\mathbf{x})$$.

is valid. with each $\mathbf{G}_i$ being a primitive $\mathscr{C'}$ mapping in some neighborhood of $0$, $\mathbf{G}_i(\mathbf{0})=0$, and $\mathbf{G'}_i(0)$ is invertible, and each $B_i$ is either a flip or the identity operator.

Here is the proof:

Put $F = F_1$. Assume $1 \leq m \leq n - 1,$ and make the following induction hypothesis ( which evidently holds for $m$= 1):

$V_m$ is a neighborhood of $0$, $F_m$ $\in$ $\mathscr C'(V_m)$, $F_m(0)$ = $0$, $F_m'(0)$ is invertible, and

$$P_{m-1}F_m(x) = P_{m-1}x ( x \in V_m). \tag{$\star$}$$

by ($\star$), we have:

$$F_m(x) = P_{m-1}x + \sum_{i=m}^n \alpha_i(x)e_i\tag{$\ast$}$$

where $\alpha_m,...,\alpha_n$ are real $\mathscr C'$-functions in $V_m$.

I don't understand from where does the last equality ($\ast$) comes.

I would be grateful for any kind of help.

Since $F$ maps $V_m$ into $\mathbb R^n,$ it has the form

$\tag 1 F_m(x)=\alpha_1(x)\vec e_1+\cdots \alpha_{m-1}(x)\vec e_{m-1}+\alpha_m(x)\vec e_m\cdots +\alpha_n(x)\vec e_n.$

Then, by definition of the projection, and by hypothesis,

$\tag 2 P_{m-1}F_{m}=\alpha_1(x)\vec e_1+\cdots \alpha_{m-1}(x)\vec e_{m-1}=P_{m-1}x.$

Now, combine $(1)$ and $(2).$