Median divisor of even perfect numbers

I noticed that when divisors of even perfect numbers are listed in ascending order, the middle divisor (I guess the median), is always of the form $$2^n$$, some power of 2. If true is there a proof for this, or does it happen all the time? I only checked up to the 8th perfect number. Thank you and apologies for the possible silliness of the question.

We know that all even perfect numbers are of the form $$(2^n-1)2^{n-1}$$ where $$2^n-1$$ is prime, so there are $$2n$$ divisors. Ignoring the divisor that is the perfect number itself (so that the remaining $$2n-1$$ divisors sum to the perfect number), it is easy to see that those divisors have a simple ordering too: $$\underbrace{1,2,\dots,2^{n-1}}_{n},2^n-1,2(2^n-1),\dots,2^{n-2}(2^n-1)$$ The median is the $$n$$th divisor in this ordering, which we see is $$2^{n-1}$$. Therefore your claim is correct.
Every even perfect number is of the form $$n=2^{p-1}(2^p-1)$$ with $$2^p-1$$ prime.
Therefore, the factors of $$n$$ are
$$1, 2, 2^2, 2^3, ..., 2^{p-1}, 2^p-1, 2(2^p-1), 2^2(2^p-1), 2^3(2^p-1), ...2^{p-2}(2^p-1),$$
and the middle one is $$2^{p-1}.$$