As others have said, the sequence is infinite, though unknowable. Even if math can't say what the precise value is, it can still prove things about the shape of the sequence.

As a simple example, we know that BB(1000) <= BB(1001) (since you can encode the longest running 1000-TM with an extra, unused state as a 1001-TM) even though we will never know either of those values.

The argument that it grows faster than any computable function is similar in that it proves things without knowing concrete numbers. (I'd have to look up the details of the proof to say more)

You should be able to prove it by showing that the existence of a faster-growing computable function lets you solve the halting problem.

Assume f(x) is computable and grows faster than BB(x). We can create a computable g(x) where g(x) > BB(x) for all x, by setting g(x) = f(x) + C for some suitable constant.

Now we create a computer program that does the following. 1. Get the size of the input program and compute g(n). 2. Run the input program for g(n) steps. 3. If it stopped, return true. If it's still running, return false.

This solves the halting problem. Since a computer program can't solve the halting problem, this program can't exist. Everything except "compute g(n)" is straightforward, so there must be no such function g, and no such function f.