Let $A_n$ denote the $n\times n$ strictly upper triangular matrix given by $$ \left( \begin{array}{cccccc} 0 & 1 & -\frac{1}{2} & & \cdots &(-1)^n\frac{1}{n-1} \\ & 0 & 1 & -\frac{1}{2} & & \\ \vdots& & 0 & \ddots & \ddots & \vdots \\ \vdots& & & \ddots & \ddots & -\frac{1}{2} \\ & & & & \ddots & 1 \\ 0 &\cdots& & & \cdots & 0 \\ \end{array} \right); $$ i.e. $A_n$ is a diagonal-constant matrix such that $[A_n]_{i,j} = (-1)^{i-j}\frac{1}{i-j}$ if $j>i$ and zero in other case.
I am stuck while trying to show that $$ \operatorname{exp}(A_n) = \operatorname{J}_n(1); $$ where $\operatorname{J}_n(1)$ is the Jordan-Block of the length $n$ for the eigenvalue $\lambda = 1$.
I have tried to give a proof by using induction without success. I will appreciate any help! Thanks.
Denote $N = J_n(1) - I_n$. We have $N^n =0$. Moreover \begin{eqnarray} A_n = N - \frac{N^2}{2} + \frac{N^3}{3}- \cdots +(-1)^{n-1}\frac{N^{n-1}}{n-1} \end{eqnarray}
Now you should check that
\begin{eqnarray} \exp\left( N - \frac{N^2}{2} + \frac{N^3}{3}- \cdots +(-1)^{n-1}\frac{N^{n-1}}{n-1}\right) = I+ N + N^n\cdot (...)= I + N \end{eqnarray}
Perhaps it is useful to check the last equality for small values of $n$.