Inspired by Jake Lai

Note that,

$\displaystyle \lim_{n \to \infty} \left(\zeta(n) - 1\right) = 0$

Consider,

$\displaystyle \lim_{n \to \infty} n(\zeta(n) - 1)$

Using L' Hopital's Rule, we have,

$\displaystyle \lim_{n \to \infty} n(\zeta(n) - 1) = \lim_{n \to \infty} \left(\dfrac{\zeta(n) - 1}{\frac{1}{n}}\right) = \lim_{n \to \infty} \left(\dfrac{\zeta '(n)}{-\frac{1}{n^2}}\right) \qquad \left( \because \dfrac{\to 0}{\to 0}\right)$

$\displaystyle \implies \lim_{n \to \infty} \left(\zeta(n) -1\right) = -n \zeta '(n)$

Now,

$\displaystyle \lim _{ n \to \infty } \dfrac { \zeta '(n+\pi )(\zeta (n)-1) }{ \zeta '(n)(\zeta (n+\pi )-1) } = \lim _{ n\to \infty } \dfrac { \zeta '(n+\pi )(n \zeta '(n)) }{ \zeta '(n)((n + \pi) \zeta '(n+\pi )) }$

$\displaystyle = \lim_{n \to \infty} \dfrac{n}{n+ \pi} = \boxed{1}$