How do I find the general term of this sequence, in terms of n?

$$\bullet$$ This is taken out of my notes and basically I have to fill in the blanks(1 and 2) but I'm stuck. Can't seem to come up with a general term by even by looking at these patterns. Would appreciate some help here!

Question:

A sequence $$u_1,u_2,u_3...$$ is given by $$u_{r+1}=\frac{r+1}{r^2}u_r$$ and $$u_1=1$$. Find an expression for $$u_n$$ in terms of n.

$$u_2=\frac{2}{1^2}$$

$$u_3=\frac{3}{2^2}\cdot\frac{2}{1^2}$$

$$u_4=\frac{4}{3^2}\cdot\frac{3}{2^2}\cdot\frac{2}{1^2}$$

1. Therefore, $$u_n=$$ ....

Alternative method:

$$u_n=\frac{(n-1)+1}{(n-1)^2}u_{n-1}$$

$$\;\;\;\;\;=\frac{n}{(n-1)^2}u_{n-1}$$

$$\;\;\;\;\;=\frac{n}{(n-1)^2}\frac{n-1}{(n-2)^2}u_{n-2}$$

$$\;\;\;\;\;=\frac{n}{(n-1)^2}\frac{n-1}{(n-2)^2}\frac{n-2}{(n-3)^2}u_{n-3}$$

1. Therefore $$u_n=$$ ....
• Is it not $\frac{n}{(n-1)!}$ Oct 30 '18 at 14:29

$$=\prod_{r=1}^m\dfrac{(n+1-r)}{(n-r)^2}u_{n-m}=u_{n-m}\dfrac n{(n-m)\prod_{r=1}^m(n-r)}$$
Set $$m=n-1$$