Why is the result of $\int_0^1 (\ln (1/x))^{p-1}dx=\Gamma (p)$?

Question

I solved this and found the answer to be $-Gamma (p)$. The book gives the answer as $Gamma (p)$. Which of us is right?

My sol:

Let $x=e^{-u}$, $dx = -e^{-u}du$, $\ln(1/x)=\ln(1)-\ln(x)$

$$\to \int u^{p-1}(-e^{-u})du$$ $$=-\Gamma (p)$$

Answer 1

Because your limits of integration will be inverted.