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+Rw=z+R$$for some constant R. Now just write down the inverse:$$z=w-Ry=v-Q(z)=v-Q(w-R)x=u-P(y,z)=u-P(v-Q(z),w-R).$$x=u-P(y,z)=u-P(v-Q(w-R),w-R).$$

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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: $$z=w-R$$ $$y=v-Q(z)=v-Q(w-R)$$ $$x=u-P(y,z)=u-P(v-Q(z),w-R).$$