Proving that $T(X)$ is a sufficient statistic for $\theta$

Question

I have that $X=(X_1,...,X_n)$ is a rv consisting of $n$ iid exponential rv's where $\theta$ is the parameter (and thus mean $1\over \theta$) .

I have to prove that $T(X)=\sum^n_{i=1}X_i$ is a sufficient statistic for $\theta$ using the theorem in my textbook which states that if $\frac{f_{\theta}(x)}{f^T_{\theta}(T(x))}$ is constant in $\theta$ then $T(X)$ is a sufficient statistic for ${\theta}$.

So what I've done is:

First I calculated $f_{\theta}(x)= \theta^n e^{-\theta\sum^n_{i=1}X_i}$ since $x=(x_1,...,x_n)$ are iid.

Then I calculated $f_{\theta}^T(T(x))= \frac{{\sum^n_{i=1}X_i}^{n-1}e^{\frac{\sum^n_{i=1}}{\theta}}}{\theta^n(n-1)!}$ since $T(X)\sim gamma(n,\theta)$

So now using the theorem with the ratio I get stuck because the $\theta$'s don't cancel out:

$$\frac{f_{\theta}(x)}{f^T_{\theta}(T(x))}=\frac{\theta^n e^{-\theta\sum^n_{i=1}X_i}}{\frac{{\sum^n_{i=1}X_i}^{n-1}e^{\frac{\sum^n_{i=1}}{\theta}}}{\theta^n(n-1)!}}$$ $$=$$ $$\frac{\theta^{2n} (n-1)!e^{\frac{1-\theta^2}{\theta}\sum^n_{i=1}x_i}}{{\sum^n_{i=1}}^{n-1}}$$

So by my calculation(probably which are wrong), $\theta$ is not cancelling out. Any mistakes that I have made that you could possibly see? Any help would be appreciated!

Here's the theorem in my textbook:

Answer 1

Because the individuals are sampled independently from an exponential distribution with parameter $\theta$, we have got $$ T(\boldsymbol{X})=\sum_{i=1}^{n} X_i \sim \Gamma(\alpha=n,\beta=\theta).$$ So, $$f_\theta^{T}(T(\boldsymbol{X}))=\theta^n \frac{1}{(n-1)!} (\sum_{i=1}^{n} X_i)^{n-1} e^{-\theta \sum_{i=1}^{n} X_i }$$ and $$ \frac{f_{\theta}(x)}{f_\theta^{T}(T(\boldsymbol{X}))} =\frac{ (n-1)!}{\sum_{i=1}^{n} X_i},$$ which is a constant function with respect to $\theta$.