Recursive sequences I have $$s_1 = 1, s_n = ns_{n-1}$$
I don't know what this means at all, sequence 1 equals 1, sequence number = number times sequence subscript number - 1
Is that it? Because it doesn't work at all when I try to work it out.
 A: I take it you mean $s_1=1, s_n=ns_{n-1}$. This is a recursive definition of the factorial, $s_n=n!$. The factorial of any natural number is that number times the factorial of its predecessor. For instance, $5!=5\cdot 4!$. Unwinding the recursion, you find that this is equivalent to saying that the factorial of a natural number is the product of all natural numbers up to and including that number. For instance,
$$
\begin{eqnarray}
5!&=&5\cdot4!
\\
&=&
5\cdot(4\cdot3!)\\
&=&
5\cdot(4\cdot(3\cdot2!))\\
&=&
5\cdot(4\cdot(3\cdot(2\cdot 1!)))\\
&=&
5\cdot(4\cdot(3\cdot(2\cdot 1)))\\
&=&
5\cdot4\cdot3\cdot2\cdot1\;.
\end{eqnarray}
$$
A: Referring to $s_1$ as "sequence one" indicates your confusion. It is more accurately referred to as the "first element of the sequence."  The sequence is the entire list of values $\{s_1,s_2,...,s_n,...\}$.
So we define the first element of the sequence, and then we define, for any $n$, the $n$th element of the sequence in terms of the $(n-1)$th term of the sequence.
A: We use $s_n=ns_{n-1}$:
$$s_1=1$$
$$s_2=2s_{2-1}=2s_1=2$$
$$s_3=3s_2=3\cdot 2=6$$
$$s_4=4s_3=4\cdot 6=4\cdot3\cdot 2=24$$
The idea is that we're given a starting number, and a procedure that will get us from one number to the next. We have a starting position and a way to move forward. This in total gives us our entire sequence. In this case it's relatively easy to see that we're multiplying all natural numbers up to $n$ (called $n!$ or "n factorial"), as joriki said. $s_n$ is not the sequence as a whole, it's the $n^{th}$ element of the sequence.
