If a program terminates and there is no one to see it, does it halt? It's time to face the truth: We will not be here forever, but at least we can write a program that will outlive the human race even if it struggles till the end of time.
Your task is to write a program that has a expected running time greater than the remaining time till the end of the universe.
You may assume that:

The universe will die from entropy in 101000 years.
Your computer:

Will outlive the universe, because it is made of Unobtainium.
Has infinite memory/stack/recursion limit.
Its processor has limited speed.



You must show that your program terminates (sorry, no infinite loops) and calculate its expected running time.
The standard loopholes apply.
This is a code golf challenge, so the shortest code satisfying the criteria wins.
EDIT:
Unfortunately, it was found (30min later) that the improbability field of Unobtainium interferes with the internal clock of the computer, rendering it useless. So time based programs halt immediately. (Who would leave a program that just waits as its living legacy, anyway?).
The computer processor is similar to the Intel i7-4578U, so one way to measure the running time is to run your program in a similar computer with a smaller input (I hope) and extrapolate its running time.

Podium
#CharsLanguageUpvotes        Author        
1    5      CJam              20       Dennis                  
2    5      J                          5         algorithmshark      
3    7      GolfScript       30        Peter Taylor          
4    9     Python                 39        xnor                        
5    10   Matlab             5          SchighSchagh      
* Upvotes on 31/08
JavaScript, 39

(function f(x){for(;x!=++x;)f(x+1)})(0)


Explanation

Since JavaScript does not precisely represent large integers, the loop for(;x!=++x;) terminates once x hits 9007199254740992.

The body of the for loop will be executed Fib(9007199254740992) - 1 times, where Fib(n) is the nth fibonacci number.

From testing, I know my computer will do less than 150,000 iterations per second. In reality, it would run much slower because the stack would grow very large.

Thus, the program will take at least (Fib(9007199254740992) - 1) / 150000 seconds to run. I have not been able to calculate Fib(9007199254740992) because it is so large, but I know that it is much larger than 101000 * 150,000.

EDIT: As noted in the comments, Fib(9007199254740992) is approximately 4.4092*101882393317509686, which is indeed large enough.