This Riemann zeta function limit is always zero for any k, right? I am quite certain that this limit:
$$\lim_{s\to 1} \, \zeta (s) \prod _{n=1}^k \left(1-\frac{1}{n^{s-1}}\right)=0$$
is always zero for any integer $k \geq 2$.
Can you prove it? I can't.
I have to go outdoors now.
As a Mathematica program this is:
Table[Limit[Zeta[s]*Product[(1 - 1/n^(s - 1)), {n, 1, k}], s -> 1], {k, 2, 12}]

 A: From $1-\frac1{1^{s-1}}=0$, the expression is zero for any $s\ne1$, hence also zero in the limit
A: To make things a bit more interesting, I will assume that you intended to start the product from $n=2$.
In this answer, I use the fact that
$$
\zeta(z)(1-2^{1-z})=1^{-z}-2^{-z}+3^{-z}-4^{-z}+\dots
$$
The idea there is to use the fact that as $z\to1$ the limit of the right hand side is $\log(2)$. Thus, we have
$$
\begin{align}
\lim_{z\to1}\zeta(z)\prod_{n=2}^k(1-n^{1-z})
&=\lim_{z\to1}\zeta(z)(1-2^{1-z})\prod_{n=3}^k(1-n^{1-z})\\
&=\log(2)\prod_{n=3}^k\lim_{z\to1}(1-n^{1-z})\\
&=0
\end{align}
$$
for $k\ge3$ since $\lim\limits_{z\to1}(1-n^{1-z})=0$ for each $n$.
A: $1-a^{1-s} = 1 - e^{(1-s)\ln a} = (s-1)\ln a + \mathcal{O}((1-s)^2)$.
that's how we can prove (*) that $\lim_{s \to 1} (1-2^{1-s}) \zeta(s) = \ln 2$.
now $1-a^{1-s}$ has a simple zero at $s=1$, thus $\prod_{n=2}^k (1-n^{1-s})$ has a $0$ of order $k-1$ at $s=1$, and $\lim_{s \to 1} \zeta(s)^{k-1} \prod_{n=2}^k (1-n^{1-s}) = \prod_{n=2}^k \ln n$ .
(*) and the fact that $\zeta(s) = \frac{s}{s-1}- s \int_1^\infty \{ x \} x^{-s-1}dx $ 
