## Task 

Using `n` as a parameter in the function `pattern`, where `n>0`, complete the codes to get the pattern (take the help of examples):

**Note:** There is no newline in the end (after the pattern ends)


### Examples

`pattern(3)` should return `"1\n1*2\n1**3"`, e.g. the following:
```
1
1*2
1**3
```
`pattern(10):` should return the following:
```
1
1*2
1**3
1***4
1****5
1*****6
1******7
1*******8
1********9
1*********10
```

Source: codewars | Difficulty: EASY
Tags: String algorithms

Fundamentals

Starter code: def pattern(n):
	

Solutions:
def pattern(n):
	return '\n'.join(['1'] + ['1' + '*' * (i - 1) + str(i) for i in range(2, n + 1)])


def pattern(n):
	s = ''
	for i in range(n):
		if i == 0:
			s += '1\n'
		else:
			s += '1{}{}\n'.format('*' * i, i + 1)
	return s.rstrip('\n')


def pattern(n):
	return '\n'.join(('{}{}{}'.format(1 if i else '', '*' * i, i + 1) for i in range(n)))


def pattern(n):
	return '1\n' + '\n'.join((f"1{'*' * (i - 1)}{i}" for i in range(2, n + 1)))


def pattern(n):
	return '\n'.join(['1' + '*' * i + str(i + 1) * (i != 0) for i in range(n)])


def pattern(n):
	if n == 1:
		return '1'
	elif n > 1:
		return pattern(n - 1) + '\n1{}{}'.format('*' * (n - 1), n)