Hypothesis testing with an exponential distribution I have the following problem:

Given the data $X_1, X_2, \ldots, X_{15}$ which we consider as a sample from a distribution with a probability density of $\exp(-(x-\theta))$ for $x\ge\theta$. 
We test the $H_0: \theta=0$ against the $H_1: \theta>0$. As test statistic $T$ we take $T = \min\{x_1, x_2, \ldots, x_{15}\}$ . Big values for $T$ indicate the $H_1$. Assume the observed value of $T$ equals $t=0.1$. 
What is the p-value of this test?
Hint: If $X_1, X_2,\ldots,X_n$ is a sample from an $\operatorname{Exp}(\lambda)$ distribution, than  $\min\{X_1,  X_2,\ldots,X_n\}$ has an $\operatorname{Exp}(n\lambda)$ distribution.

The solution says 0.22. 
I know that the first question you have to ask youself regarding the p-value is: 
"What is the probability that the H0 would generate a sample θ>0?"
So I assume H0 is true and take θ = 0. The probability-density function becomes:
f(x) = Exp(-x). I take up the hint, so I make it f(x) = Exp(-nx)
This is where I get stuck. I don't know how to proceed with the information given:

Assume the observed value of T equals t=0.1.

Can I have feedback on this problem? 
Thanks,
Ter

You know that the test statistic under the null hypothesis has distribution
$$T\sim \operatorname{Exp}(n)=\operatorname{Exp}(15)$$
The weight of the tail of this distribution is
$$\operatorname{P}(T>t)=\exp(-15t)$$
We reject null at significance level
$$\alpha \leq \operatorname{P}(T>t)=\exp(-15t)$$
p-value is maximum significance level at which we reject null, so
$$\text{p-value}=\exp(-15t)=\exp(-15\times 0.1)\approx 0.22$$
