Simplification of a pairing
Surely bilinearity helps and and you have to be aware of the fact that $e(g^a,g^b)\cdot e(g^a,g^{-b})=e(g^a,g^b)\cdot e(g^a,g^b)^{-1}=1$. 

You see that from your last equation
$$=e(g^{a\lambda_i}\cdot H(x)^{-v_ir_i},g^{u})\cdot e(g^{r_i}, g^{u't}\cdot H(x)^{v_i(u+\gamma)})\cdot e(g^{r_i}, g^{-tu'})\cdot e(g, H(x)^{v_i\gamma r_i})^{-1}$$

you get 

$$=e(g^{a\lambda_i},g^u)\cdot e(H(x)^{-v_ir_i},g^{u})\cdot e(g^{r_i}, g^{u't})\cdot e(g^{r_i},H(x)^{v_iu})\cdot e(g^{r_i},H(x)^{v_i\gamma})\cdot e(g^{r_i}, g^{-tu'})\cdot e(g, H(x)^{v_i\gamma r_i})^{-1}$$

and by rearanging the terms a bit and using bilinearity, the fact that you have a symmetric pairing (switch arguments) and the fact mentioned above you will notice that only the first term remains and that is your desired result.