= Solution
The displayed matrix is an <upper bidiagonal matrix>. For $\lambda_1=1$, the <right eigenvector> equation $(A-I)v=0$ successively gives
$$
v_2=v_3=\cdots=v_n=0.
$$
We may therefore take the unit right eigenvector to be $v=e_1$.
For a <left eigenvector>, $(A^T-I)u=0$ gives
$$
u_{i-1}+(\lambda_i-1)u_i=0
\qquad(2\leq i\leq n).
$$
Since $\lambda_i-1=-1/i$,
$$
u_i=i\,u_{i-1}.
$$
Choosing the initial scale $u_1=1$ produces
$$
u=(1,2!,3!,\ldots,n!)^T.
$$
After normalization,
$$
\widehat u=
\frac{(1,2!,3!,\ldots,n!)^T}
{\sqrt{1+(2!)^2+\cdots+(n!)^2}}.
$$
The <eigenvalue sensitivity> is therefore
$$
s(1)=\frac1{|\widehat u^Tv|}
=\sqrt{1+(2!)^2+\cdots+(n!)^2}
\geq n!,
$$
where $n!$ is the <factorial>. Hence
$$
\boxed{s(\lambda_1)\geq n!}.
$$
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