Jacobian Conjecture for unit triangular matrices This question is about the Jacobian conjecture for a special case.  I will first explain the Jacobian conjecture (since it is something every mathematician should know about).
Let $k$ be an algebraically closed field.
Consider a map $$F: k^n \rightarrow k^n,$$
defined by $$F(x_1,\ldots,x_n)=(f_1(x_1,\ldots,x_n),\ldots,f_n(x_1,\ldots,x_n)),$$
where $f_1,\ldots,f_n$ are polynomials.
The Jacobian of $F$, which I denote $J$, is the determinant of the matrix $dF$ where the $(i,j)$-th entry of $dF$ is $\partial f_i/\partial x_j$.  (The matrix $dF$ gives the induced map on the tangent bundle, or maybe it's the cotangent bundle; it doesn't matter for this question.)
Since $F$ is given by polynomials, the entries of $dF$ are polynomials and $J$ is a polynomial.  Hence $F$ is a nonsingular map if and only if $J$ is a constant.
The inverse function theorem tells us that $F$ has a smooth inverse map if and only if $J$ is constant.  The Jacobian conjecture says that this smooth map is in fact also given by polynomials (in the case where the original map $F$ is given by polynomials).
Question: I would like to know if the Jacobian conjecture is known (or trivial) for the special case where the matrix $dF$ is a triangular matrix with $1$'s on the diagonal.  If it is not known, I would like to know if the full Jacobian conjecture is known to be equivalent to this special case.
Motivation: This is a possible strategy for proving that a particular family of maps I have constructed for a particular purpose is in fact invertible within the category of affine algebraic varieties.
EDIT: Clarifying in light of Tom's remark.  The inverse function theorem just says that $F$ has a local inverse.  The Jacobian conjecture is that $F$ has a global inverse which is given by polynomials.
 A: 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: 
$$z=w-R$$
$$y=v-Q(z)=v-Q(w-R)$$
$$x=u-P(y,z)=u-P(v-Q(w-R),w-R).$$
A: The Jacobian conjecture is trivial in this case.
Write $M$ for $dF$ and $N$ for its inverse, which is also triangular with 1's on the diagonal.  (I assume upper triangular.)
Now just look at the consequences of $MN=I$.
First, for each $j$, we have $M_{j,j+1}+N_{j,j+1}=0$.  This shows that every entry of the form $N_{j,j+1}$ is a polynomial.
Next, for each $j,t$ with $t>0$ we have $\sum_p M_{j,j+p}N_{j+p,j+t}=0$. This plus induction shows that $N_{j,j+t}$ is a polynomial.  (The induction is on the difference between the two indices on the entry of $N$.)
Edit:  As I said in comments, this never actually uses the upper triangularity and doesn't establish the existence of a global inverse.
Here, I think, is a much better answer:
Write $F=(F_1,\ldots,F_n)$.  We want to know whether $F$ is globally injective.
Suppose $F(s_1,\ldots,s_n)=F(t_1,\ldots,t_n)$.  We know that $F_n$ is an affine function of the last variable only, so $s_n=t_n$.  We know that $F_{n-j}$ is of the form $Ax_{n-j}+G(x_{n-j+1},\ldots,x_n)$, so by backward induction on $n$ we know that $s_{n-j}=t_{n-j}$ for all $j$.  So $F$ is globally injective, which is to say that the Jacobian conjecture holds.
