You know that \(2^{k}k! < (2k)!\) and you need to show that \(2^{k+1} (k+1)! < (2(k+1))!\)
Do a compare and contrast between corresponding parts of your known information and your goal. That is, compare \(2^{k}k!\) to \(2^{k+1} (k+1)!\) and compare \((2k)!\) to \((2(k+1))!\). In each case, can you express the larger quantity as the smaller quantity plus some extra factors?
Now, write down just the parts that are different in the two equations. Why are the extra factors in \((2(k+1))!\) larger than the extra factors in \(2^{k+1} (k+1)!\)?
Remember that k isn't just some random integer. We have a lower bound on its size. Look back at the outline if you don't remember what the bound was.
If that isn't working, double-check how you expressed \((2(k+1))!\) in terms of \((2k)!\). How many extra factors does \((2(k+1))!\) have?