I am trying to prepare for my exams and want to calculate some matrix exponents $\exp(A)$ and wanted to do this with $$A = \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}.$$
Right now I have $\sum_{j=0}^\infty \frac{(-1)^{(j-1)}}{(j-1)!}$
$$ e^{A} = \begin{bmatrix} 1/e & \sum_{j=0}^\infty \frac{(-1)^{(j-1)}}{(j-1)!} & 1 \\ 1 & 1/e & 1 \\ 1 & 1 & 1/e \end{bmatrix}.$$
I would greatly appreciate any tips or tricks I could use to progress further or even understand were I might have made mistakes.
Further info: I arrived at my question through the following:
$A = (D+N)$
$$ (D+N)^n=\sum_{n=0}^{\infty}{n \choose k} D^{n-k} \cdot N^k = {n \choose 0} D^{n} \cdot N^0 + {n \choose 1} \cdot D^{n-1} \cdot N =D^n+n \cdot D^{n-1}\cdot N$$ Since $N^n = 0$ $\forall n\geq 2$ only the first two sum elements matter.
Rewrite the sum as $$\sum_{j=0}^\infty \frac{(-1)^{(j-1)}}{(j-1)!}\overset{k=j-1\ }{\longrightarrow}\sum_{k=-1}^\infty \frac{(-1)^{k}}{k!}=-\frac{1}{(-1)!}+\sum_{k=0}^\infty \frac{(-1)^{k}}{k!}.$$
In order to understand what $\frac{1}{(-1)!}$ means, we have to look at the reciprocal Gamma function. We can express our term like this: $\frac{1}{(-1)!}=\frac{1}{\Gamma(0)}=0.$ If the starting point is actually $j=1$ and not $j=0$ we get the same result, since both outcomes are zero and do not change the value of the series.
Lastly, note that for all $x\in\mathbb R$, $\mathrm e^x = \sum_{n=0}^\infty \frac{x^k}{k!}$, so our final solution is $e^{-1}$.