The answer by Bazin (https://mathoverflow.net/users/21907/bazin), Faa di Bruno's formula for vector valued functions, URL (version: 2012-09-04): https://mathoverflow.net/q/106339
is providing a formula for the $n$th derivative of $f\circ g$. Assuming that $g=\Phi$ is invertible and smooth, we take $f=\Phi^{-1}$ its inverse function and we find for $n\ge 2$
$$
0=\sum_{\substack{n_1+\dots+n_r=n\\r\ge 1, n_j\ge 1}}
\frac{f^{(r)}\circ g}{r!}\frac{g^{(n_1)}}{n_1!}\dots \frac{g^{(n_r)}}{n_r!},
$$
so that writing the "first" term for $r=n, n_j=1$, we get 
$$
\frac{f^{(n)}\circ g}{n!}\bigl(g^{(1)}\bigr)^n+\sum_{\substack{n_1+\dots+n_r=n\\n>r\ge 1, n_j\ge 1}}
\frac{f^{(r)}\circ g}{r!}\frac{g^{(n_1)}}{n_1!}\dots \frac{g^{(n_r)}}{n_r!}=0,
$$
and thus an explicit formula expressing  $$
\bigl[(\Phi^{-1})^{(n)}\circ \Phi\bigr]\times \frac{(\Phi^{(1)})^n}{n!}
$$
as a linear combination of the $r$th derivative of $\Phi^{-1}$ (with $1\le r<n$)
whose coefficients are products $\frac{\Phi^{(n_1)}}{n_1!}\dots \frac{\Phi^{(n_r)}}{n_r!}$ where $n_1+\dots+n_r=n$.
We get that 
$$
\underbrace{[(\Phi^{-1})^{(n)}\circ \Phi]\times \frac{(\Phi^{(1)})^n}{n!}}_{\Psi_n}
=-\sum_{\substack{n_1+\dots+n_r=n\\n>r\ge 1, n_j\ge 1}}
\underbrace{{[(\Phi^{-1})^{(r)}\circ \Phi]}
\times \frac{(\Phi^{(1)})^r}{r!}}_{=\Psi_r}
 {(\Phi^{(1)})^{-r}}
\frac{\Phi^{(n_1)}}{n_1!}\dots \frac{\Phi^{(n_r)}}{n_r!}
,
$$
that is 
$$
\Psi_n=-\sum_{n>r\ge 1}
\Psi_r
 \underbrace{
{(\Phi^{(1)})^{-r}}
\sum_{\substack{n_1+\dots+n_r=n\\ n_j\ge 1}}
\frac{\Phi^{(n_1)}}{n_1!}\dots \frac{\Phi^{(n_r)}}{n_r!}}_{\Omega_r(\Phi)},\quad
\text{and we have also }
\Psi_1=I,
$$ 
so that
$$
\Psi_n=-\sum_{1\le r\le n-1}\Psi_r \Omega_r(\Phi), \quad \Psi_1=
I.$$
We can now prove inductively that $\Vert\Psi_n\Vert$ is bounded above by a  polynomial $$
P_n(\frac{\Vert \Phi^{(\nu_1)}\Vert}{\Vert \Phi^{(1)}\Vert },
\dots,
\frac{\Vert \Phi^{(\nu_l)}\Vert}{\Vert \Phi^{(1)}\Vert }), \quad \nu_j\le n.
$$