Python Permutation Backtracking - Why do I have to call list in this backtracking example?

Question

In the code blow why do I need to call list() in the base case? If I don't call the list on current_perm, it just returns a list of empty lists.

        permutations = []

# returns a list of its permutation
        def permutation(nums, current_perm = []):
            if len(nums) == 0:
                # must call list on current_perm
                # not sure why
                permutations.append(list(current_perm))
            else:
                for idx, num in enumerate(nums):
                    current_perm.append(num)
                    # pass nums as nums without the current number
                    permutation(nums[:idx]+nums[idx+1:], current_perm)
                    # backtrack
                    current_perm.pop()