You've gotten to the step where you assume p(k)

2^(k+2)>= 2k + 5 

It's a good idea to figure out what you want to prove first
before proving it.

What you want to prove is p(k+1), which would be

2^(k+1+2) >= 2(k+1) + 5

2^(k+3) >= 2k + 2 + 5

2*(k+3) >= 2k + 7

That's what we want to prove, just don't forget that we haven't
proved it yet!

Then we ask ourselves "How do I go from the left side of
p(k) to the left sides of p(k+1)"

That is:

How do I go from 2^(k+2) to 2^(k+3)?  The answer is to multiply
by 2.  So we multiply both sides of p(k) by 2, and hope that
we get something greater or equal to the right side of p(k+1)
which is 2k + 7.

2[2^(k+2)]>= 2[2k + 5]

2^(k+3) >= 4k + 10

Now all you need do is show that 4k + 10 is greater than or equal to 
2k + 7 which is the right side of p(k+1).

And certainly since k is a positive integer 4k >= 2k and 10 >= 7,
and adding those inequalities give 4k + 10 >= 2k + 7

So

2^(k+3) >= 4k + 10 >= 2k + 7

Edwin