= Solution
The <Kolchin theorem> conjugates a faithful representation of a <unipotent algebraic group> $U$ into the upper unitriangular group $U_n$. Let $U_n^{(r)}$ consist of matrices whose first $r-1$ superdiagonals vanish. Matrix multiplication gives
$$
[U_n^{(r)},U_n^{(s)}]\subseteq U_n^{(r+s)}.
$$
Since $U_n^{(n)}=1$, this filtration is a finite <central series>, so $U_n$ and every subgroup of it are <nilpotent groups>. Hence $U$ is nilpotent.
The converse fails: $\mathbb G_m$ is abelian, hence nilpotent, but its nonidentity points are semisimple rather than unipotent.
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