По данному натуральному n ≤ 9 выведите лесенку из n ступенек

n = int(input())
 
for i in range(n):
    for i in range(1, i+2):
        print(i, end='')

n = 9
for i in range(n):
    print(''.join(map(str, range(1, i+2))))

n = 9
print("\n".join(''.join(map(str, range(1, i+2))) for i in range(n)))

n = int(input())
 
for i in range(n):
    for j in range(1, i+2):
        print(j, end='')
    print()

r = '123456789'
for i in xrange(n+1):
    print r[:i]

print """
1
12
123
1234
12345
123456
1234567
12345678
123456789
"""

s = """1
12
123
1234
12345
123456
1234567
12345678
123456789
"""
s1 = s[:n*(n+3)/2 ]
print s1

n=9
print("""
1
12
123
1234
12345
123456
1234567
12345678
123456789
"""[:(n+3)*n//2+1])

class spruce:
    s = '123456789'
 
    def __init__(self, n):
        self.nu = n
 
    def get_spruce(self):
        for i in range(self.nu + 1):
            print(self.s[:i])
 
 
z = spruce(9)
z.get_spruce()

import numpy as np
buf = np.zeros((n, n+1), dtype=np.byte)
arr = np.arange(ord('1'), ord(str(n+1))+1, dtype=np.byte)
buf[:] = arr
#print buf
ig, jg = np.mgrid[0:n,0:n+1]
#print jg > ig
arr2 = np.where(jg > ig, ord(' '), buf)
arr2.dtype = 'S1'
arr2[:, n] = '\n'
print arr2.tostring()

arr = np.arange(ord('1'), ord('1') + n+1, dtype=np.byte)

a=int(input())
s=""
for i in range (a):
    s=s+str(a-a+i+1)
    print(s)

import numpy as np
n=7
buf = np.tril(np.cumsum(np.ones((n, n), dtype=np.uint8), axis=1))
buf2='\n'.join(''.join(str(cell) for cell in row) for row in buf)
print(buf2.replace('0', ''))