Python 2, 59 55 53 bytes

lambda s:s in[len(s)/i*s[:i]for i in range(1,len(s))]

Try it online!

Boring regex version:

Python 2, 44 bytes

lambda s:re.match(r'(.+)\1+$',s)>0
import re

Try it online!

Python 2, 24 bytes

lambda s:s in(s*2)[1:-1]

Try it online!

Shamelessly stolen from xnor's answer to the original question.

More intuitive version:

Python 2, 59 55 53 bytes

lambda s:s in[len(s)/i*s[:i]for i in range(1,len(s))]

Try it online!

Boring regex version:

Python 2, 44 bytes

lambda s:re.match(r'(.+)\1+$',s)>0
import re

Try it online!

Python 2, 59 55 53 bytes

lambda s:s in[len(s)/i*s[:i]for i in range(1,len(s))]

Try it online!

Recursive version:

Python 2, 59 54 bytes

f=lambda s,i=1:s[i+1:]and f(s,i+1)or s==len(s)/i*s[:i]

Try it online!

Boring regex version:

Python 2, 44 bytes

lambda s:re.match(r'(.+)\1+$',s)>0
import re

Try it online!

Python 2, 59 55 53 bytes

lambda s:s in[len(s)/i*s[:i]for i in range(1,len(s))]

Try it online!

Boring regex version:

Python 2, 44 bytes

lambda s:re.match(r'(.+)\1+$',s)>0
import re

Try it online!

Python 2, 59 bytes

lambda s:any(s.replace(s[:i],'')==''for i in range(len(s)))

Try it online!

Recursive version:

Python 2, 59 bytes

f=lambda s,i=0:s[i+1:]and f(s,i+1)or''==s.replace(s[:i],'')

Try it online!

Boring regex version:

Python 2, 44 bytes

lambda s:re.match(r'(.+)\1+$',s)>0
import re

Try it online!

Python 2, 59 55 53 bytes

lambda s:s in[len(s)/i*s[:i]for i in range(1,len(s))]

Try it online!

Recursive version:

Python 2, 59 54 bytes

f=lambda s,i=1:s[i+1:]and f(s,i+1)or s==len(s)/i*s[:i]

Try it online!

Boring regex version:

Python 2, 44 bytes

lambda s:re.match(r'(.+)\1+$',s)>0
import re

Try it online!