Q:

Is there any efficient way to compute the exponentiation of a fraction and an integer?

Mind the following JavaScript operation:
function exp(n){
    return Math.floor(Math.pow(1.01, n));
}

Where r is a fraction (such as 1.01) and n is an integer. Is there any efficient way to emulate that operation efficiently on Ethereum? That is, a function:
function exp(uint n) returns (uint) {
    return ...
}

Which returns similar values to the former? Such a function would have many use cases; for example, computing compound interest rates.

A:

Approximate solution using the binomial expansion
The following is a decent, low-cost approximation:
// Computes `k * (1+1/q) ^ N`, with precision `p`. The higher
// the precision, the higher the gas cost. It should be
// something around the log of `n`. When `p == n`, the
// precision is absolute (sans possible integer overflows). <edit: NOT true, see comments>
// Much smaller values are sufficient to get a great approximation.
function fracExp(uint k, uint q, uint n, uint p) returns (uint) {
  uint s = 0;
  uint N = 1;
  uint B = 1;
  for (uint i = 0; i < p; ++i){
    s += k * N / B / (q**i);
    N  = N * (n-i);
    B  = B * (i+1);
  }
  return s;
}

This function computes k * (1+1/q) ^ N. So, for example, to compute 2500 * 1.01 ^ 137, we could use fracExp(2500, 100, 137, 8). This outputs 9769, which is very close to the real value, 9771.657061221898.

Explanation
1. Representing rational numbers with fixed points
The main issue is Ethereum doesn't have floating points. The usual approach to represent real numbers is by using fixed points. That means we multiply everything by a large number and assume there is an implicit dot. For example, 1.002 is represented as 1002. For most arithmetic operations we don't really need floats; examples:

Addition: 1.1 + 1.2 == 11/10 + 12/10 == 23/10 == 2.3
Multiplication: 1.1 * 2 == 11/10 * 2 = 22/10 == 2.2

2. Fixed point exponentiation
For exponentiation, though, that approach doesn't work so well. To implement it, we'd need to exponentiate the numerator and the denominator, and then divide. Example: 1.1^75 == (11/10)^75 == (11^75)/(10^75) == 1271. The problem is that, to get to the result, we needed to compute the intermediate value 11^75, which is larger than 2^256, resulting in an integer overflow. Both operands and the results could fit in a 16-bit uint, yet we overflow the 256-bit word size! 
A way to avoid the overflow is to implement exponentiation as a loop; at each pass, you multiply the number and re-normalize it so it doesn't get too big. The issue with that approach is that it is linear on the exponent, so it might be prohibitely expensive. There are more efficient exp implementations such as exponentiation by squaring; but I'm not sure that approach allows normalizing the intermediate terms.
3. Avoiding big intermediate values with binomial expansion
Let's look again at an example of the kind computation we're trying to perform:
1.01 ^ 100

We can rewrite this as:
(0.01 + 1) ^ 100

Applying the binomial expansion, that becomes:
  1                  * 1 ^ 100 * 0.1 ^ 0
+ 100                * 1 ^  99 * 0.1 ^ 1
+ 100*99/2           * 1 ^  98 * 0.1 ^ 2
+ 100*99*98/2/3      * 1 ^  97 * 0.1 ^ 3
+ 100*99*98*97/2/3/4 * 1 ^  96 * 0.1 ^ 4
...
+ 1                  * 1 ^   0 * 0.1 ^ 100

Since 1^N is always 1, this is the same as:
  1                  * 0.1 ^ 0
+ 100                * 0.1 ^ 1
+ 100*99/2           * 0.1 ^ 2
+ 100*99*98/2/3      * 0.1 ^ 3
+ 100*99*98*97/2/3/4 * 0.1 ^ 4
...
+ 1                  * 0.1 ^ 100

Notice that each new line contains progressively smaller terms. Due to that, we're able to eliminate a good chunk of those lines without affecting the result considerably. That solution gives us a reasonable approximation. The solution posted here computes the p first lines of that series.

