This is my solution. First, represent the $$$n$$$ numbers in binary. Each number has $$$k$$$ digits, where each digit can be $$$0$$$ or $$$1.$$$ Let $$$a_{i, j}$$$ represent the $$$j$$$th digit (from left to right) of $$$a_i$$$, where $$$a_i$$$ is represented in binary.
Now,
is true if for all $$$1 \le j \le k,$$$
Let's consider the inequality for $$$j = 1,$$$ because others values of $$$j$$$ are similar.
If more than one of $$$a_{i,1} = 0$$$ for $$$1 \le i \le n,$$$ then the left side of the above inequality is going to be zero. This requires the right side of the above inequality to be zero, so an even number of $$$a_{i,1}$$$ should be $$$1$$$.
Also, if all of $$$a_{i,1}=1,$$$ then that is another possible case. This only needs to be considered if $$$n$$$ is odd.
Now, using the above two paragraphs, there are
ways to pick the $$$1$$$st digit for all $$$a_i.$$$
Using the Binomial Theorem on $$$(1+1)^n$$$ and $$$(1-1)^n,$$$ we can show that $$$\sum_{p=0}^{\lfloor n/2 \rfloor} \binom{n}{2p} = 2^{n - 1}.$$$
Thus, the number of ways to pick the $$$1$$$st digit for all $$$a_i$$$ is $$$2^{n - 1} + n\mod 2.$$$ Similarly, this is the number of ways to pick the $$$j$$$th digit for all $$$a_i.$$$
Since there are $$$k$$$ digits total in each of $$$a_i,$$$ the answer is $$$(2^{n - 1} + n \mod 2)^k.$$$
My code describes this approach, and I think it performs this approach correctly. Thus, I believe my approach to the problem is incorrect. Can anyone point out what is wrong?
I have also attached by code in case it is wrong.
Thanks for the help everyone!!