An example of what I was talking about as far as the exponentiation thing is concerned
Consider an operator that functions as such, let us call it R for reduce, as the operator is called in programming. R takes a function and a list. The function specifies how the values of the list will be combined in the following manner. Consider f(x, y) R of the list L, in this example this is how the result is defined f(f(f(...f(f(f(first l, second l), third l), fourth l), (l's size - 1) l), last l). Another function is then defined as the map function M, this function is defined as taking a function and a list as well and specifies the values of the list will be transformed. Consider a M applied to list L with function f(x) in the form f(x) M of the list L, here the result is the list {f(first l), f(second l), f(third l)...f((l's size-1) l), f(last l)}. We also define the operator TO which takes two numbers and constructs a list containing them. For example take the numbers n and m, the result will be the list {n, n+1, n+2...m-1, m}
We can then define summation from k = n to m of f(k) as x + y R (f M (n TO m)). We can also define product from k = n to m of f(k) as x * y R (f M (n TO m)). I will now define the RR, or repeat reduce, operator as an operator that takes a function f and applies it to create f R (g M (n TO m)) where n, m, and g are passed into the function.
We now define a new operator E, the exponentiation operator from k = n to m of f(k) as x^y R (f M (n TO m)). This operator takes on a couple of special properties. One of the more useful properties is that it can be expressed in terms of product. The equation for this is such, E from k = n to m of f(k) = f(n) ^ (the product from k = n + 1 to m of f(k)). This follows from the standard exponentiation rules. We can then make a few interesting equations, for one any exponentiation where f(n) = 0 gives 1. We can also say the following, the exponentiation from k = 1 to infinity of x^k = x^(the product from k = 2 to infinity of x^k) = x^((the product from k = 1 to infinity of x^k)/x) = x^(x^(-1/12)/x) = x^x^(-13/12)
the product from k = 1 to infinity of x^k = x^(-1/12) is able to be constructed from the summation of natural numbers being -1/12. as that product is just equal to x^(the summation from k = 1 to infinity of k)
); myself and 
