Number of FLOPs (floating point operations) for exponentiation
Assuming multiplication between two numbers use one FLOP, the number of operations for $x^n$ will be $n-1$. However, is there a faster way to do this ...


There most certainly is a faster way to do this for non-negative integer powers. For example, $x^{14}=x^{8}x^{4}x^{2}$. It takes one multiplication to compute $x^2$, one more to compute $x^4$, one more to compute $x^8$, and two more to multiple those three numbers. This suggests a simple cost and a simple algorithm.


Convert the non-negative integer power to base 2.
Count the number of ones in this representation.
Add the power of two corresponding to the most significant non-zero bit in this representation.
Subtract one.


This yields a concise algorithm for any non-negative integer power. This algorithm is the most efficient, up to $x^{14}$. This algorithm suggests six multiplications are needed to compute $x^{15}$ since $x^{15}=x^8x^4x^2x$. However, 15 is 120 in base 3 and 30 in base 5, both of which imply that only five multiplications are needed to compute $x^{15}$: $x^{15}=(x^3)^4x^3$ from the base three representation, and $x^{15}=(x^5)^3$ from the base five representation. The minimum number of multiplications needed to compute $x^n$ where $n$ is a non-negative integer is in fact an NP-complete problem. But it's a whole lot less than $n-1$ multiplications.


  ... and how does it work if $n$ isn't an integer?


There are some tricks one can use if $n$ is a rational. But if $x$ is real and $n$ is a non-negative real, one must resort to approximation techniques. (For example, approximation techniques are used twice-fold in calculating $\exp(n\ln(x))$.)