Computing a certain $2014$-fold product using a particular associative binary operation $\ast$ 
$$x*y = 3xy - 3x - 3y + 4$$
We know that $*$ is associative and has neutral element, $e$.
Find $$\frac{1}{1017}*\frac{2}{1017}*\cdots *\frac{2014}{1017}.$$

I did find that $e=\frac{4}{3}$, and, indeed, $x*y = 3(x-1)(y-1)+1$. Also,it is easy to check that the law $*$ is commutative.
How can I solve this?
 A: Hint What is $1 \ast y$?

Additional hint Note that $1$ occurs in the $2014$-fold product, namely as $\frac{1017}{1017}$.

A: One easy way to manufacture non-obvious associative/commutative operations is to take a known associative operation and conjugate it with an invertible function $f : \mathbb R \to \mathbb R$.  In this case, taking $f(x) = 3(x-1)$ (with inverse function $f^{-1}(x) = 1 + \tfrac x3$), we can see that
$$f(x*y) = 9(x-1)(y-1) = f(x)f(y),$$
so that the mysterious $*$ operation is just multiplication conjugated by $f$, i.e. $x*y = f^{-1}(f(x)f(y))$.  This makes it very clear why it is both associative and commutative: it mimics those properties from multiplication.  Also note that $e = f^{-1}(1)$, with $1$ being the neutral element for multiplication.
Just as diagonalizing a matrix $A$ into $PDP^{-1}$ lets you compute powers easily, conjugation lets you easily express longer chains:
$$f(x*(y*z)) = f(x)f(y*z) = f(x)f(y)f(z).$$
More generally, $f(a_1 * a_2 * \cdots * a_n) = \prod_{i=1}^n f(a_i)$, so we can compute
$$a_1 * a_2 * \cdots * a_n = f^{-1}\left(\prod_{i=1}^n f(a_i)\right) = 1 + \tfrac13 \prod_{i=1}^n 3(a_i-1) = 1 + 3^{n-1} \prod_{i=1}^n (a_i-1).$$
It so happens that for your question, one of the $a_i$s is equal to $1$, simplifying the whole expression.  But even without this great simplification, it would be easy to compute $a_1 * a_2 * \cdots * a_n$.
