Question Statement
A matrix $N \in M_{n}(\mathbb{R})$ is called nilpotent if there exist $k$ such that $N^{k} = 0 \in M_{n}(\mathbb{R})$. The smallest such $k$ is called the index of $N$. Let $$ A = \lambda I + N$$
where $\lambda \in \mathbb{R}$ and $N$ is nilpotent of index $k$. Obtain a simple (that is easy to compute ) formula for the matrix exponential $e^{At}$.
Attempt at solution \begin{align*} e^{At} &= \sum_{k=0}^{\infty} \frac{t^k}{k!}(\lambda I +N)^{k} \\ &=\sum_{k=0}^{\infty} \frac{t^k}{k!} \sum_{n=0}^{k-1} \binom{k}{n} (\lambda I)^{k-n} N^{n} \\ &=\sum_{n=0}^{k-1}\sum_{k=0}^{\infty} \frac{k!}{n! (k-n)!} \lambda^{k-n} N^{n} \\ &=\sum_{n=0}^{k-1}\sum_{k=0}^{\infty} \frac{t^{k-n}t^{n}}{(k-n)!n!} \lambda^{k-n}N^{n} \\ &\text{Letting $ m = k -n $ we get } \\ &=\sum_{n=0}^{m=0}\sum_{k=0}^{\infty} \frac{t^m}{m!} \lambda^{m} \frac{N^n t^n}{n!} = \sum_{n=0}^{k-1} e^{\lambda t} \frac{N^{n}t^n}{n!} \end{align*} I don't know how to proceed from here or if my solution makes sense in the first place.