Why print(5 ** 2 ** 0 ** 1) = 5 in Python?

Because, like many languages, the exponentiation operator binds right-to-left:

5 ** 2 ** (0 ** 1)
==
5 ** (2 ** 0)
==
5 ** 1
==
5

Exponentiation is right-associative, so your expression is the same as

5 ** (2 ** (0 ** 1))
   == 5 ** (2 ** 0)
   == 5 ** 1
   == 5

where any integer raised to the zeroth power is 1 by definition.

Others have already pointed this out already, but I just wanted to mention the documentation. You can see this by typing help("OPERATORS") in your repl. There you will spot somewhere at the top:

Operators in the same box group left to right (except for exponentiation, which groups from right to left).

You are right to be surprised though, this seems like a very odd decision to me. In other languages, e.g. octave, 5 ^ 2 ^ 0 ^ 1 == 1 as you'd expect. Oddly enough, both Julia and R agree with python on this.

EDIT: On second thought, I suppose making exponentiation right-associative makes sense too; you would expect to mean 5^8 rather than 25^3 ...

Incidentally, here's another trap. How much is: 5 * -1 ** 2 ? Is it 5 or -5?

See help("**") to see why it is what it is. (incidentally, this is how octave, R, and julia treat this case too).

The moral of the story is: always group when there is potential for ambiguity. Or at least check the precedence is what you think it is before ungrouping.

** is exponentiation.

0 raised to the power of 1 is 0. So, we could re-write the statement as print(5**2**0) without changing the result.

2 raised to the power of 0 is 1. So, we can re-write the statement as print(5**1) without changing the result.

5 raised to the power of 1 is 5. So, we can rewrite the statement as print(5) without changing the result.