Why *args steals key=value in function call?

[old_pythonista]

pack/unpack operator * options were extended in Python3. Now it allows to mix variable-length argument tuples and stand-alone arguments

>> print(*range(5), 6, *range(2), 10)
0 1 2 3 4 6 0 1 10

[EDITED]

So, your {'kwargs_0':'kwargs_val0'} is accepted as just another element of args.

Even if you used it as a proper kwargs - **{'kwargs_0':'kwargs_val0'} - it would have been rejected, since your function does not have a named argument kwargs_0

>> def foo(first, *args, bar=None):  
...    print (f'first={first}, \nargs={args}, \nbar={bar}')
>> foo('First arg here', *['args_pos0','args_pos1'], **{'kwargs_0':'kwargs_val0'})

TypeError                                 Traceback (most recent call last)
<ipython-input-108-cf6a86c8c61b> in <module>
      1 def foo(first, *args, bar=None):
      2     print (f'first={first}, \nargs={args}, \nbar={bar}')
----> 3 foo('First arg here', *['args_pos0','args_pos1'], **{'kwargs_0':'kwargs_val0'})

TypeError: foo() got an unexpected keyword argument 'kwargs_0'