By Proposition 3.24, we write

(3.278)		$\displaystyle\psi$	$\displaystyle=\FT_{1}+\FT_{2}+\FT_{3},$
(3.279)		$\displaystyle\FT_{1}$	$\displaystyle\equiv\frac{\sqrt{-1}}{4r}dy\wedge d\bar{y},$
(3.280)		$\displaystyle\FT_{2}$	$\displaystyle\equiv\frac{1}{2r}(yd\bar{y}+\bar{y}dy)\wedge\Gamma=O^{\prime}(1),$
(3.281)		$\displaystyle\FT_{3}$	$\displaystyle\equiv r\cdot d\Gamma+r^{-3}\Pi_{2}^{(4)}+O^{\prime}(r^{2})=O^{\prime}(r).$

Immediately we have $(\FT_{1})^{2}=(\FT_{2})^{2}=\FT_{1}\wedge\FT_{2}=0$ and for all $k\geq 2$, $\FT_{2}\wedge(\FT_{3})^{k}=(\FT_{3})^{k}=O^{\prime}(r^{k})$. Moreover,

(3.282)

$$\FT_{1}\wedge\FT_{3}=\frac{\sqrt{-1}}{4}dy\wedge d\bar{y}\wedge d\Gamma+\frac{\sqrt{-1}}{4r^{4}}dy\wedge d\bar{y}\wedge\Pi_{2}^{(4)}+O^{\prime}(r).$$

Notice that $\frac{\sqrt{-1}}{4}dy\wedge d\bar{y}\wedge d\Gamma$ is a smooth term and by definition $dy\wedge d\bar{y}\wedge\Pi_{2}^{(4)}=0$, then

(3.283)

$$\FT_{1}\wedge\FT_{3}=O^{\prime}(r).$$

So for all $k\geq 1$,

(3.284)

$$\FT_{1}\wedge(\FT_{3})^{k}=O^{\prime}(r^{k}).$$

Now by direct calculation,

(3.285)

$\displaystyle\psi^{k}=k\Big((\FT_{1})\wedge(\FT_{3})^{k-1}+(\FT_{2})\wedge(\FT_{3})^{k-1}\Big)+(\FT_{3})^{k}.$

The conclusion then follows. ∎