Minimal polynomial of matrix exponential

Question

Let $A$ be a $n\times n$ matrix with minimal polynomial $m_A(t)=t^n$, i.e. a matrix with $0$ in the main diagonal and $1$ in the diagonal above the main diagonal.

How can I show that the minimal polynomial of $e^A$ is $m_{e^A}(t)=(t-1)^n$? I have already calculated that $e^A$ is a matrix with $1$ in the main diagonal, $1$ in the diagonal above the main diagonal, $\frac{1}{2}$ in 2nd diagonal above the main diagonal and so forth until $\frac{1}{(n-1)!}$ in the upper right entry. But there must be some easier way to justify $(t-1)^n$ as the minimal polynomial besides brute force calculation?

Answer 1

Let $n\ge2.$ Clearly, $$e^A=\sum_{k=0}^{n-1}{1\over k!}A^k\tag1$$ By (1), for a positive integer $r,$ $$(e^A-I)^r=\left(\sum_{k=1}^{n-1}{1\over k!}A^k\right)^r\tag2$$

Let $(f(x))^r$ be the polynomial obtained from the right side of (2) by replacing $A$ by the variable $x.$ Since $f(x)$ has the form $\ x + x^2g(x)\ $ when $n\ge3,$ $$(f(x))^r=x^r\left(1+xg(x)\right)^r\tag3$$ By (3), $n$ is the least value of $r$ such that $x^n$ divides $(f(x))^r.$ Therefore, $n$ is the least value of $r$ such that both sides of (2) are zero. Thus, $m_{e^A}(t)=(t-1)^n.$

Answer 2

so $A = N$ and $e^A = I + N + \frac{1}{2!}N^2 +...+\frac{1}{(n-1)!}N^{(n-1)}$
$A$ having minimal polynomial $x^{n}$ tells you that
$\big\{I,N,N^2,...N^{n-1}\big\}$ are linearly independent

since $e^A$ is upper triangular with ones on the diagonal all eigenvalues are 1 and $(x-1)^n$ annihilates $e^A$ per Cayley Hamilton.

It is enough to show that nilpotent $B:= e^A-I=N + \frac{1}{2!}N^2 +...+\frac{1}{(n-1)!}N^{(n-1)}$ is not annihiled by a polynomial of degree $n-1$

Since the minimal polynomial for $B$ divides its characteritic polynomial, it suffices to show that $B^{n-1}\neq \mathbf 0$
$B^{n-1} $
$= \Big(N + \frac{1}{2!}N^2 +...+\frac{1}{(n-1)!}N^{(n-1)}\Big)^{n-1}= \Big(N + \big(\frac{1}{2!}N^2 +...+\frac{1}{(n-1)!}N^{(n-1)}\Big)\Big)^{n-1} = N^{n-1} + \mathbf 0$
$\neq \mathbf 0$