show/hide this revision's text 2 added 17 characters in body

More simply:

(I'll write this down for the case $n=3$ because writing and reading subscripts makes me tired.)

Let $(u,v,w)=F(x,y,z)$. By hypothesis $u-x$ has derivative $0$ with respect to $x$, so $u=x+P(y,z)$ $u=x+P(y,z)$$ for some $P$. And $v-y$ has derivative $0$ with respect to $x$ and $y$, so $v=y+Q(z)$ $v=y+Q(z)$$ for some $Q$. And $w-z$ has derivative $0$ with respect to all three variables, so $w=z+R$ $w=z+R$$ for some constant $R$.

Now just write down the inverse: $$z=w-R$$ $$y=v-Q(z)=v-Q(w-R)$$ $$x=u-P(y,z)=u-P(v-Q(z),w-R).$$$x=u-P(y,z)=u-P(v-Q(w-R),w-R).$$

show/hide this revision's text 1

More simply:

(I'll write this down for the case $n=3$ because writing and reading subscripts makes me tired.)

Let $(u,v,w)=F(x,y,z)$. By hypothesis $u-x$ has derivative $0$ with respect to $x$, so $u=x+P(y,z)$ for some $P$. And $v-y$ has derivative $0$ with respect to $x$ and $y$, so $v=y+Q(z)$ for some $Q$. And $w-z$ has derivative $0$ with respect to all three variables, so $w=z+R$ for some constant $R$.

Now just write down the inverse: $$z=w-R$$ $$y=v-Q(z)=v-Q(w-R)$$ $$x=u-P(y,z)=u-P(v-Q(z),w-R).$$