Not considering number 1, the integer 153 is the first integer having this property:
the sum of the third-power of each of its digits  is equal to 153. Take a look:
153 = 1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153

The next number that experiments this particular behaviour is 370 with the same power.

Write the function `eq_sum_powdig()`, that finds the numbers below a given upper limit `hMax` that fulfill this property but with different exponents as a power for the digits.

eq_sum_powdig(hMax, exp): ----> sequence of numbers (sorted list) (Number one should not be considered).

Let's see some cases:
```python
eq_sum_powdig(100, 2) ----> []

eq_sum_powdig(1000, 2) ----> []

eq_sum_powdig(200, 3) ----> [153]

eq_sum_powdig(370, 3) ----> [153, 370]
```
Enjoy it !!

Difficulty: introductory

Starter code:

def eq_sum_powdig(hMax, exp):
	

Solutions:
def eq_sum_powdig(hMax, exp):
    return [i for i in range(2, hMax + 1) if i == sum(int(c) ** exp for c in str(i))]


def eq_sum_powdig(hMax, exp):
    ret = []
    for i in range(10, hMax + 1):
        if i == sum(int(j) ** exp for j in str(i)):
            ret.append(i)
    return ret

def eq_sum_powdig(hmax, exp):
    return [i for i in range(2, hmax+1) if sum(int(d) ** exp for d in str(i)) == i]

def valid(num, power):
    return num == sum(int(x)**power for x in str(num))
def eq_sum_powdig(hMax, exp):
    return [x for x in range(2,hMax+1) if valid(x,exp)]

def eq_sum_powdig(hMax, exp):
    list=[]
    for n in range(100,hMax+1):
        product=0
        for l in str(n):
            product += int(l)**exp
        if n == product:
            list.append(n)
    return list