If $U$ is a closed subgroup of $GL(V)$ consisting of unipotent elements, show that $\log x$ $(x \in U)$ belongs to $\mathfrak{u}$.
Here, for an unipotent element $x \in GL(V)$, $\log x = \sum_{i=1}^{\infty}(-1)^{i+1}(x-1)^i/ i$. $\mathfrak{u}$ is the Lie algebra of $U$. All can be represented in matrix form.
If I am not mistaken, $x$ can be transformed into a triangular matrix with diagonal elements all equal $1$. So $T_{ij}$ $(1 \leq j < i \leq n)$, $T_{kk}$ $(1 \leq k \leq n)$ are all in $\mathscr{I}(U)$, the defining ideal of $U$. Suppose that $\mathrm{dim}V = n$, then every element of $y \in U$, satisfies $(y -1)^n =1$. Then how can I find a polynomial $f(T_{11}, \cdots, T_{nn}) \in \mathscr{I}(U)$, according to the equality $(y -1)^n =1$? I think this might be helpful for the proof.
Then, how can I prove it?