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!