Use the Erlang :math module

:math.pow(2,3) #=> 8.0

If you want an integer:

:math.pow(2,3) |> round #=> 8

Erlang's :math.pow has some limitations, for example it will not allow really high integer exponents:

iex(10)> :math.pow(2, 10000)
** (ArithmeticError) bad argument in arithmetic expression

You can easily reimplement a fast algorithm for computing exponentials that will work with the arbitrarily large integers provided by the runtime:

defmodule Pow do
require Integer


def pow(_, 0), do: 1
def pow(x, n) when Integer.is_odd(n), do: x * pow(x, n - 1)
def pow(x, n) do
result = pow(x, div(n, 2))
result * result
end
end


iex(9)> Pow.pow(2, 10000)
19950631168807583848837421626835850838234968318861924548520089498529438830...

Here is a tail call optimized implementation of the power function:

def  pow(n, k), do: pow(n, k, 1)
defp pow(_, 0, acc), do: acc
defp pow(n, k, acc), do: pow(n, k - 1, n * acc)

Tis works - it will be great when I learn enough to know exactly why it works - probably something to do with eval under the covers:

defmodule Example do
require Integer


def do_it(list) do
list
|> Enum.reject(&Integer.is_odd(&1))
|> Enum.map(&(:math.pow(&1,3)))
end


end

If the base is 2 and the power is an integer, you can do a left bitshift using the function Bitwise.bsl. For example, 2³ can be calculated with:

> Bitwise.bsl(1, 3)
8

Integer.pow/2

As of v1.12, Integer.pow/2 is the way to do this.

**/2

As of Elixir 1.13, ** is available.

> 2 ** 3
8

Note: it returns a float if the exponent is less than 0.

Documentation