# Inequality Induction problem 1

### Hints for inductive step

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?