Let X1, · · · , Xn be an i.i.d. sample from an exponential distribution with the density function. Find the maximum likelihood of $$f(x|β) = \frac{1}{β}\,e^{-x/β}$$
First, I need to find: $$p(x|β) = \prod_{i=1}^n \frac{1}{β} \,e^{-x_i/β}$$ but I'm not sure how I can simplify the equation above.
Here's my attempt: $$p(x|β) = \frac{1}{β^n} \,e^\frac{-\sum_{i=1}^nx_i}{β}$$
Then, I have to find the negative log-likelihood: $$L(x|β) = \ nlog(β) + \frac{\sum_{i=1}^nx_i}{β}$$
Then, I have to differentiate with respect to β to get the MLE. So $$β = \frac{\sum_{i=1}^nx_i}{n}$$
Is this correct? I'm unsure about my simplification.