Before we start our new python project here is another solution for one of the python’s question on

def isPP(power): for num in range(1, power): for po in range(2, power): if(num ** po == power): return [num, po] return None

The above method is only suitable to solve a number input which is below 1800, the program will become freeze if the number goes beyond that value (depends on the processing power of your own computer). Well, hope you like this quick solution, in the next chapter we will begin our project.

Instead, find the prime factors of the number.

Stop when you find some prime factors that show that we can’t make a**b -> n.

https://gist.github.com/gcapell/4309fdb0ee281e06bb71a9b694d81711

Finds the answer for 1800 **14 in fraction of a second.

no need to return None, it does that by default.