I saw the following formula (in (7.25) of Lie Groups, Physics, and Geometry)
$$ \exp\left( \begin{array}{c} w & z\\ 0 & 1\\ \end{array} \right) = \left( \begin{array}{ccc} e^w & (e^w-1)z/w\\ 0 & 1 \\ \end{array} \right). $$
Is this correct or some kind of typo? At least for $z=0$, it does not seem right?
The typo is in the last element of left-hand side matrix. The correct equation would be: $$ \exp\left( \begin{array}{c} w & z\\ 0 & 0\\ \end{array} \right) = \left( \begin{array}{ccc} e^w & (e^w-1)z/w\\ 0 & 1 \\ \end{array} \right) $$
The formula is indeed incorrect, including the case when $z=0$.
In the simple case that $z=0$, you have a diagonal matrix, and the exponential of a diagonal matrix $\exp\begin{pmatrix}a&0\\0&b\end{pmatrix}$ is the diagonal of the exponentials $\begin{pmatrix}e^a&0\\0&e^b\end{pmatrix}$. In your case it would give $$\exp\begin{pmatrix}w&0\\0&1\end{pmatrix}=\begin{pmatrix}e^w&0\\0&e\end{pmatrix}.\tag1$$
In the general case that $z\neq0$, you need a general expression for the $n$th power of you matrix $M=\begin{pmatrix}w&z\\0&1\end{pmatrix}$, which is easy to see is of the form $M^n=\begin{pmatrix}w^n&u_n\\0&1\end{pmatrix}$, where $u_1=z$ and $u_{n+1}=wu_n+z$, so therefore $u_{n+1}=z\sum_{k=0}^nw^k=z\frac{w^{n+1}-1^{n+1}}{w-1}$.
Since the matrix exponential is defined as $\exp M=\sum_{n=0}^\infty\frac1{n!} M^n$, everything stays just as in (1) except the right-hand corner, which is given by $$ \left(\exp M\right)_{12}=\sum_{n=0}^\infty\frac{u_n}{n!} =\frac{z}{w-1}\left[\sum_{n=0}^\infty\frac{w^n}{n!}-\sum_{n=0}^\infty\frac{1^n}{n!}\right] =z\frac{e^w-e}{w-1}. $$
The final result is then $$\boxed{\exp\begin{pmatrix}w&z\\0&1\end{pmatrix}=\begin{pmatrix}e^w&z\frac{e^w-e}{w-1}\\0&e\end{pmatrix}.}\tag2$$
