Avoiding Python tricks to find anagrams
Taking advantage of the work previously submitted by @hjpotter92 and adding an extra twist:
def anagram_check(*args ):
    return len( { tuple(sorted(x.lower().replace(' ', ''))) for x in args } ) == 1

This, potentially, gains an additional pythonicity score by avoiding repetition of the sorted(x.lower().replace( ...) construct. Instead, the construct is applied to both arguments in the set comprehension, and the test to see if the resulting set has a single element provides the answer. This implementation also generalizes the API to cover more that two strings if wanted, returning True if all strings given are anagrams of each other (thanks to Heap Overflow for pointing this out).
If you really want a function which is limited to two strings, you might prefer:
def anagram_check(string1, string2):
    return len({tuple(sorted(x.lower().replace(' ', ''))) for x in [string1,string2]}) == 1

The code can be made less terse and more legible by defining a function to act on each string:
def make_unique(x): 
     return tuple( sorted( x.lower().replace(' ', '') ) )

def anagram_check(*args): 
     return len( { make_unique(x) for x in args } ) == 1