An infinite product function

Before I start, must be said that I am not a math wizard, nor a math student. I just love nudging around with math, and I came across a random function I thought in my head (I do not take credit for nothing, what I mean by 'I though'=what came to my mind) So sorry if the maths you see here are corrupt, and sorry if I post about something that was discovered long time ago.

The function is $$ P(k) = \prod_{i=2}^{\infty} \left( 1- \frac{1}{i^k} \right )$$

For any $k \in (0, \infty)$. I tried playing around with different values of $k$ and go these results:

$$\begin{array}{|c|c|}\hline k& P(k) \\ \hline 2 & 0.5 \\ \hline 3 & 0.80939 \\ \hline 4 & 0.91901 \\ \hline 5 & 0.96325\\ \hline \end{array}$$

Now, of-course this is not 'that' interesting for $k \in \mathbb{Z}$ because it tends to $1$ pretty quickly, even from $k \le 9$ we are close to $1$ by a per-mille - $P(9) = 0.99799$. And so I wanted to see what happens for values $k = [0, 10]$ using the numpy library to use linspace as so:

space = np.linspace(0, 10, 100000) # This is our range

Which splits the intervals $[0,10]$ into 10K values (each are equally far away from each-other) - then calculated $P(k) ~~ \forall k \in \text{space}$ and got this beautiful output:

It look roughly like what I expected it to look like, with an asymptote $y=1$ as we tend to $1$ as $k$ grows. Meaning our function (not specifically the one in the picture) is defined as so:

$$ P : [0, \infty) \to [0, 1) \\ P(k) = \prod_{i=2}^{\infty} \left( 1- \frac{1}{i^k} \right ) $$

We can calculate $P(2)$ very easily as so:

$$ p(2) := \prod_{i=2}^{\infty} \left( 1- \frac{1}{i^2} \right) =\prod_{i=2}^{\infty} \left( \frac{i^2 - 1}{i^2} \right) = \prod_{i=2}^{\infty} \left( \frac{(i+1)(i-1)}{i^2} \right) ~~ \text{\\} \lg$$

$$\begin{aligned} \lg(p(2)) &= \sum_{i=2}^{\infty} \lg \left( \frac{(i+1)(i-1)}{i^2} \right) = \lim_{N \to \infty} \sum_{i=2}^{N} \lg \left( (i+1)(i-1) \right) - 2\lg (i) \\&= \lim_{N \to \infty} \lg \left ( \frac{(N+1)!}{2} \cdot (N-1)! \cdot \frac{1}{(N!)^2} \right) = \lg(0.5) = \lg(P(2)) \implies P(2) = \frac{1}{2} \end{aligned}$$

As for any other $k$, it might be impossible to solve it as we did for $P(2)$ because $\prod \left (i^3 - 1 \right)$ does not factor nicely.

The Python Code (Uses numpy and matplotlib):

import numpy as npimport matplotlib.pyplot as pltspace = np.linspace(0, 10, 10000)Y = np.array([0])X = spacefor x in X[1:]:    a = 1    for i in range(2, 1000):        a *= (1 - 1/(i ** x))    Y = np.append(Y, [a])plt.scatter(X, Y)plt.show()

The for-loop with the range (2, 1000) tries to approximate $\prod_{i=2}^{\infty}$ (And of-course I don't have enough 'time'😉 to wait until $\infty$.. so I chose a big enough value that would output a good approximation).

My questions are a bit "dry" when it comes to the mathematics behind this function.

1. Is this kind of function known somewhere? Used in any way?
2. The function (for me) looks like it has a logarithmic style, is it related in any way? any approximations can be done to this function? (As I said, I am not a math-wiz unfortunately..)

Thank you!