How to show that $$\left\lfloor\zeta\left(1+\frac{1}{n}\right)\right\rfloor=n$$ for every positive integer $n$?



Floor of Riemann zeta function
It is known that (see Corollary 1.14 in Montgomery-Vaughan: Multiplicative number theory I)
$$\frac{1}{\sigma-1}<\zeta(\sigma)<\frac{\sigma}{\sigma-1},\qquad \sigma\in(0,1)\cup(1,\infty).$$
In particular, taking $\sigma=1+\frac{1}{n}$, we get
$$n<\zeta\left(1+\frac{1}{n}\right)<n+1.$$
This is slightly stronger than your claim. Better bounds can be obtained from the Laurent series expansion of $\zeta(s)$ around $s=1$.


