Nilpotent matrices and unipotent matrices

Finite sums. If $N^n=0$ then $N^k=0$ for all $k\ge n$, so
$$e^N=\sum_{k=0}^{n-1}\frac{N^k}{k!},\qquad\log(I+N)=\sum_{k=1}^{n-1}\frac{(-1)^{k+1}}kN^k.$$
Both are polynomials in $N$ (of degree $<n$), with fixed rational coefficients, so as maps of the entries of $N$ they are polynomial, hence continuous.

Images. $e^N-I=N\bigl(I+\frac N2+\cdots\bigr)=N\cdot P(N)$ where $P(N)$ commutes with $N$, so $(e^N-I)^n=N^nP(N)^n=0$: $e^N\in\mathcal{U}$. Similarly $\log(I+N)=N\cdot Q(N)$ is nilpotent ($N$ and $Q(N)$ commute). So $\exp:\mathcal{N}\to\mathcal{U}$ and $\log:\mathcal{U}\to\mathcal{N}$ are well defined.

Mutually inverse. The formal identities $\exp(\log(1+x))=1+x$ and $\log(\exp x)=x$ hold in $\mathbb{Q}[[x]]$. Substituting a nilpotent $N$ with $N^n=0$ amounts to working modulo $x^n$ (a finite computation), so the identities hold for $N$:
$$e^{\log(I+N)}=I+N,\qquad\log(e^N)=N.$$

Conclusion. Both $\exp|_{\mathcal{N}}$ and $\log|_{\mathcal{U}}$ are polynomial in the entries, inverse to each other. So $\exp$ is a homeomorphism (indeed a polynomial bijection with polynomial inverse).

Example. For $N=\begin{pmatrix}0&a\\0&0\end{pmatrix}$: $e^N=\begin{pmatrix}1&a\\0&1\end{pmatrix}$ and $\log\begin{pmatrix}1&a\\0&1\end{pmatrix}=N$.

Remark. This fails for non-unipotent matrices, where $\log$ is only defined locally (see the previous problem) and $\exp$ is not even injective.