```
power :: (a -> a) -> Int -> a -> a
power f 0 = (\x -> x)
power f n = (\x -> f (power f (n-1) x))
```

Since the function is called `power`, there is one obvious application that springs to mind, i.e. raising to a power:

```
raise :: Int -> Int -> Int
raise x y = power (*x) y 1
```

## 2 responses to “Exercise 9.8”

The following is a possible definition for power using

higher-order functions:

It can be simplified to:

power f n = foldr (.) id (replicate n f)