The expression is ill defined for $x>1$ (except for integer $x$), because then $(1-x)/x < 0$.
In this situation, one usually defines
$$a^b = \exp\big(b\,\text{Log}(a)\big)$$
where $\text{Log}$ is a complex logarithm.
In case you're not familiar with complex logarithms, here's a summary. $\exp$ is periodic in the complex plane: we have $\exp(2\pi i) = 1$, and hence $\exp(z + 2k\pi i) = \exp(z)$ for all $k\in\Bbb Z$.
This means that the complex exponential does not admit a classical inverse, so the complex logarithm is 'multivalued'.
Indeed, we have
$$\exp^{-1}\left(re^{i\theta}\right) = \{\log(r)\,+i(\theta + 2k\pi)\mid k\in\Bbb Z\}.$$
When doing calculations with the complex logarithm, one chooses a branch.
Usually, the branch cut is done along the negative real axis; this is called the principal branch of the complex logarithm.
That your expression is ill defined it easy to see even if we ignore the underlying nuances of complex numbers.
In general, if $a<0$ and $q \in \Bbb Q\setminus\Bbb Z$, then $a^q$ is not well defined in the reals.
Say $q = m/n$ with $m,n\in\Bbb Z$.
Is $a^q = \sqrt[n]{a^m}$, or is it ${\left(\sqrt[n]a\right)}^m$?
Notice that if $n$ is even, then $\sqrt[n]a$ is not well defined, so the order matters (but ideally it shouldn't).
If we choose to define $a^q = \sqrt[n]{a^m}$, when $m$ is odd then $a^m<0$, so for even $n$ the root is still ill-defined.
And even if $n$ were odd, we still run into problems:
if $q=m/n$, then $q=(km)/(kn)$ for any $k\in\Bbb Z$.
In particular, we may choose $k$ even.
The value or 'well-definedness' of $a^q$ must not depend on a particular representation of $q$.
And after all this, there's the problem of defining $a^b$ for $b>0$ irrational.
This is usually done for $a>0$ via continuity: we show that $a^q$ is continuous on the positive rationals, and define
$$a^b = \lim_{\substack{q\,\in\,\Bbb Q\\q\,\to\,b}} a^q.$$
But if $a<0$, $a^q$ is ill defined on the rationals, so how do you define $a^b$?
If you require $x\in\Bbb N$ as $x\to\infty$, then we have
$${\left(\frac{1-x}x\right)}^x = {(-1)}^x\,{\left(1 - \frac1x\right)}^x. \tag{$*$}$$
A classical limit is $\lim_{n\to\infty}{\left(1 + \frac1n\right)}^n = e$.
Indeed, one often defines $e$ this way.
Now, consider
\begin{align}
{\left(1 + \frac1n\right)}^n \cdot {\left(1 - \frac1n\right)}^n
&=
{\left(1 - \frac1{n^2}\right)}^n \leqslant 1.
\end{align}
By Bernoulli's inequality, we have
$$
{\left(1 - \frac1{n^2}\right)}^n \geqslant 1-\frac1n,
$$
so by the squeeze theorem $\lim_{n\to\infty} {\left(1 + \frac1n\right)}^n \cdot {\left(1 - \frac1n\right)}^n = 1$.
It follows from the algebra of limits that $\lim_{n\to\infty} {\left(1 - \frac1n\right)}^n$ exists and
$$\lim_{n\to\infty} {\left(1 - \frac1n\right)}^n
=
\frac{1}{\lim_{n\to\infty} {\left(1 + \frac1n\right)}^n}
= \frac1e.$$
Hence, even for integer $x$, the expression $(*)$ fails to have a limit: for even $x$, $(*)$ approaches $1/e$, while for odd $x$ it approaches $-1/e$.
If $x$ is a rational multiple of $\pi$, then for some integer $N > 0$, $\sin(Nx) = 0$. This forces $\prod_{n=1}^r \sin(nx) = 0$ whenever $r \ge N$. In this case, the limit is $0$.
Otherwise, $x$ is not a rational multiple of $\pi$ and $|\cos x| < 1$. Notice
$$|\sin(nx)\sin(n+1)x| = \frac{|\cos x - \cos((2n+1)x)|}{2} \le \mu \stackrel{def}{=}\frac{1 + |\cos x|}{2}$$
By grouping the factors in the numerator in pairs, we have following bound
for the weighted product at finite $r$.
$$r\left|\prod_{n=1}^r \sin(nx)\right|
\le r\prod_{k=1}^{\lfloor r/2\rfloor} |\sin((2k-1)x)\sin(2kx)|
\le r\mu^{\lfloor r/2\rfloor}
$$
Since $\mu < 1$, the limit of the weighted product is again $0$.
Best Answer
Consider the sequence $x_n$ defined by $x_0 = x$ and $x_{n+1} = \cos(x_n)$. Then the sequence in question is $$a_n = n\prod_{n=1}^\infty x_n.$$ I claim that $a_n \to 0$. Here's a sketch of the proof:
The hard bit is 2, which I'll leave to last. To prove 1, note that the function $f(x) = x - \cos(x)$ is continuous, negative at $0$, and positive at $\pi/2$, so by the intermediate value theorem, there must be at least one point where $f(x) = 0$ in $[0, \pi/2]$.
Further, $f'(x) = 1 + \sin(x) \ge 0$, meaning that the function is non-decreasing. If $f$ had more than one root, then it'd be an interval of roots, which would correspond to an interval of roots in $f'$. This is clearly not the case, so there is a unique $x^*$ such that $f(x^*) = 0$, i.e. $\cos(x^*) = x^*$.
The point $x^*$ lies in the range of $\cos$, i.e. $[-1, 1]$, as well as $[0, \pi/2]$, so $x^* \in [0, 1]$. If $x^* = 1$, then $\cos(x^*) = 1$, hence $x^*$ would have to be an integer multiple of $2\pi$, which it is clearly not. Thus, $x^* \in [0, 1)$, as claimed.
To prove 3, assuming 2 is proven, consider $$\left|\frac{a_{n+1}}{a_n}\right| = \frac{n+1}{n}|x_{n+1}| \to x^* < 1,$$ thus the series converges absolutely. Then, 4 follows immediately from this: the terms of a convergent series must tend to $0$.
Now, we tackle 2. First, recall the trigonometric identity: $$\cos(x) - \cos(y) = -2\sin\left(\frac{x + y}{2}\right)\sin\left(\frac{x - y}{2}\right).$$ Now, suppose that $x, y \in [0, 1]$. Note that $\frac{x + y}{2} \in [0, 1]$ and $\sin$ is increasing and positive on $[0, 1] \subseteq [0, \pi/2]$, hence $$\left|\sin\left(\frac{x + y}{2}\right)\right| = \sin\left(\frac{x + y}{2}\right) \le \sin(1).$$ Also, recall that $|\sin \theta| \le |\theta|$ for all $\theta$. Hence, assuming still $x, y \in [0, 1]$, $$|\cos(x) - \cos(y)| = 2\left|\sin\left(\frac{x + y}{2}\right)\sin\left(\frac{x - y}{2}\right)\right| < 2 \cdot \sin(1) \cdot \left| \frac{x - y}{2}\right| = \sin(1)|x - y|.$$
Now, note that $x_n \in [-1, 1]$ for $n \ge 1$, and since $\cos$ is positive over $[-1, 1]$, we have $x_n \in [0, 1]$ for $n \ge 2$. So, for $n \ge 2$, we get $$|x_{n+2} - x_{n+1}| = |\cos(x_{n+1}) - \cos(x_n)| \le \sin(1)|x_{n+1} - x_n|.$$ This implies the series $$\sum_{n=2}^\infty (x_{n+1} - x_n)$$ is absolutely summable, as it passes the ratio test (limsup version): $$\left|\frac{x_{n+2} - x_{n+1}}{x_{n+1} - x_n}\right| = \frac{|x_{n+2} - x_{n+1}|}{|x_{n+1} - x_n|} \le \sin(1) \frac{|x_{n+1} - x_n|}{|x_{n+1} - x_n|} = \sin(1) < 1.$$ Therefore, $\sum_{n=2}^\infty (x_{n+1} - x_n)$ converges. This is a telescoping series, whose partial sums take the form $x_n - x_2$. These partial sums converge, and hence so must $x_n$.
Now, because $x_n$ converges to some $L$, it follows from $\cos$ being continuous that $$x_{n+1} = \cos(x_n) \implies L = \cos(L) \implies L = x^*,$$ completing step 2, and the full proof, as necessary.