"i don't know what it's there for"
[tex]![/tex] is the factorial symbol, and it's used to denote the product of successively smaller non-negative integers. That is, if [tex]n=10[/tex], for instance, then [tex]n![/tex] is the product of all positive integers less than or equal to 10:
[tex]10!=10\cdot9\cdot8\cdot\ldots\cdot3\cdot2\cdot1[/tex]
By convention, we define [tex]0!=1[/tex].
One way of interpreting the factorial is combinatorically, meaning we can use it to count the number of ways of doing some particular task. Suppose I'm getting dressed, and I have 3 basic items that I have to put on: a shirt, a pair of pants, and a pair of socks. How many ways are there in which I can put each thing on in order?
For the first item I have 3 choices (shirt, pants, or socks). Once I put on the first article of clothing, I have 2 choices left. After that, 1 choice is left. In total, I would have [tex]3\cdot2\cdot1=6[/tex] possible ways of putting on all my clothes in a particular order.
The factorial generalizes this and makes writing the product more compact: [tex]3!=3\cdot2\cdot1=6[/tex].
Back to your actual question: We're given some function [tex]P(k)[/tex], and we're asked to find the value of [tex]P(k+1)[/tex]. This is just a matter of replacing [tex]k[/tex] with [tex]k+1[/tex]:
[tex]P(k+1)=\dfrac{2^{(k+1)-1}}{(k+1)!}=\dfrac{2^k}{(k+1)!}[/tex]
and we're done.
