Given is a random sample $X_1 ... X_n$ from a $Ber(p)$ distribution. Consider the estimator $T = min\{X_1 ... X_n\}$. First, what is now the distribution of $T$?
The minimum says that everything should be working ($X=1$), so I think $T $~ $Ber(p^n)$. Is this correct?
Furthering, I would like to show that: $MSE(T) = p^n - 2p^{n+1} + p^2$. Since $T$ is not an unbiased estimator for $p$: $MSE(T) = Var(T) - (E[T]-p)^2$. So we also have to know the variance. Can somebody help me out of this?
You are right, $T$ has Bernoulli distribution parameter $p^n$. It follows that the variance of $T$ is $p^n(1-p^n)$. We can now compute $E[(T-p)^2]$ using your procedure.
Or else, perhaps a little more simply, we can expand $(T-p)^2$, getting $T^2-2pT+p^2)$. The expectation of this is $E[T^2]-2pE[T]+p^2$. Now $E[T^2]=E[T]=p^n$ since $T^2=T$. It follows that $$E[(T-p)^2]=p^n-2p^{n+1}+p^2.$$
Random variable $T$ has a Bernoulli distribution with success probability $p^n$. So as Andre said the Variance of $T$ is simply $p^n(1-p^n)$.
One calculates the Mean squared error as follows:
$$MSE=Variance+Bias^2$$ so $$MSE=p^n(1-p^n)+(E[T]-p)^2=p^n(1-p^n)+(p^n-p)^2=p^n-2p^{n+1}+p^2$$
