Truncated exponential map from $\mathfrak{gl}_n$ to $GL_n$ Let $k$ be a field of characteristic $p>0$.  If $A$ is a nilpotent matrix in $\mathfrak{gl}_n(k)$, with $p>n$, then we can define the unipotent matrix:
$$\exp(A)=\sum_{i=0}^{p-1}\frac{A^i}{i!}$$
If $f_p(x,y)=\frac{1}{p}((x+y)^p-x^p-y^p)\in k[x,y]$, and if $AB=BA$, we have the formula 
$$\exp(A)\exp(B)=\exp(A+B-f_p(A,B))\tag{$*$}$$
Notice the expansion of $f_p$ is well-defined in $k[x,y]$.  I would like to be able to prove the following:

If $A_1,\ldots,A_k$ are linearly independent, pairwise commuting, nilpotent matrices, then $\prod_{i=1}^k\exp(A_i)^{m_i}=I_n$ if and only if $m_i=0$ for $i=1,2,\ldots,k$.

One step I've made is to see that $f_p(x,x)=0$ implies that $\exp(A)^m=\exp(mA)$.  As I try to proceed further, the $f_p(x,y)$ terms get messy.
 A: Yes, this is true, assuming that by $m_i=0$ you mean $m_i \equiv 0 \mod p$. 
If the $A_i$ commute, then $\prod \exp(A_i)^{m_i} = \exp \left( \sum m_i A_i \right)$. Since the $A_i$ are commuting and nilpotent, the sum $\sum m_i A_i$ is nilpotent. The exponential map is injective on nilpotent matrices (with inverse $\log$) so $\exp \left(  m_i A_i \right)=I_n$ implies that $\sum m_i A_i=0$. Since the $A_i$ are linear independent, this means that all the $m_i$ are $0$.

In the above, I am using the following Lemma: If $p > n$, and $A$ and $B$ are commuting $n \times n$ nilpotent matrices, then $\exp(A+B) = \exp(A) \exp(B)$. The key to this is to recall that, if $A$ and $B$ are commuting nilpotent matrices, then we can choose a basis in which they are both strictly upper triangular. Using this, if $A$ and $B$ are commuting nilpotent matrices, then $A^i B^j=0$ for any $(i,j)$ with $i+j \geq n$. Thus
$$\exp(A+B) = \sum_{m < n} \frac{(A+B)^m}{m!} = \sum_{i+j < n} \frac{A^i B^j}{i! j!} = \sum_{i < n} \sum_{j < n} \frac{A^i}{i!} \frac{B^j}{j!} = \exp(A) \exp(B).$$

It's worth pointing out that this is not true if you only assume that $A_i^p=0$ and not $p>n$. For example, take $p=2$ and 
$$A = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} \quad B = \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \quad C = \begin{pmatrix} 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$
If I haven't made any errors, then $A^2=B^2=C^2=0$, $A$, $B$ and $C$ commute and $A$, $B$ and $C$ are linearly independent, but $\exp(A) \exp(B) \exp(C) = (1+A)(1+B)(1+C) = 1$. I can explain how I found this if there is a need.
