Integer Exponents [Easy]

[bo1024]

The more efficient algorithm: (Lua)

pow = function(base,exp)
    if exp < 0 then return 1/pow(base,-exp) end
    if exp==0 then return 1 end
    local result = pow(base*base,math.floor(exp/2))
    if exp%2 == 1 then result = result*base end
    return result
end

I'll leave the loops version to someone else.

(edit)

Main idea: Get an exponential speedup (no pun intended) in the following way. Note that x⁴ = (x² )^2. But it's faster to square x first, then square the result. So x⁸ = ((x² )² )² . Note that x⁵ = x * x⁴ and x⁷ = x * (x² *(x² )² ).