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Proof of inconsistency of $(n+1)X_{(1)}$
I am try to proof that $(n+1)X_{(1)}$  is inconsistent.
$F(X_{(1)}) = P(X_{(1)}<x) = P(min(X_1, ..., X_n) < x) = 1 - P(min > x) = 1 - P(X_1>x)P(X_2>x)...P(X_n>x) = 1 - P(X>x)^n$
Know that, $F(X_{(1)}) = 1 - P(X>x)^n = 1 - (\frac{\theta-x}{\theta})^n$, than
$F(X_{(1)}) = 1 - P(X>x)^n = 1 - (\frac{\theta-x}{\theta})^n$
Defenition of inconsistency:
$ \forall \epsilon>0 \lim_{n\rightarrow \inf}P(|(n+1)X_{(1)} - \theta| > \epsilon) = 0$
and from here I can't achieve inconsistency. Can anybody help? Thanks in andvanced

Note that $(n+1)X_{(1)}$ is consistent for $\theta$ iff for any $\epsilon>0$,
$$
\lim_{n \to \infty} P(|(n+1)X_{(1)} - \theta| < \epsilon) = 1.
$$
Since $F_{X_{(1)}}(x)=1-(1- \frac{x}{\theta})^n$, we have
\begin{align*}
P(|(n+1)X_{(1)} - \theta| < \epsilon) 
&= P( \theta-\epsilon < (n+1) X_{(1)} < \theta+\epsilon)\\
&= P( \frac{\theta - \epsilon}{n+1} <  X_{(1)} < \frac{\theta+\epsilon}{n+1})\\
&=\left[  
1-\left ( 1-\frac{\theta + \epsilon}{\theta(n+1)}\right)^n
\right]-
\left[  
1-\left ( 1-\frac{\theta - \epsilon}{\theta(n+1)}\right)^n
\right]\\
&=  
\left ( 1-\frac{\theta - \epsilon}{\theta(n+1)}\right)^n
-
\left ( 1-\frac{\theta + \epsilon}{\theta(n+1)}\right)^n
\\
&=  
\left ( 1-\frac{1-\epsilon/\theta}{n+1}\right)^n
-
\left ( 1-\frac{1+\epsilon/\theta}{n+1}\right)^n
\end{align*}
Therefore
$$
\lim_{n \to \infty} P(|(n+1)X_{(1)} - \theta| < \epsilon)
= e^{1-\epsilon/\theta}- e^{1+\epsilon/\theta}
\neq 1
$$

Sorry, but didn't you forget about $\theta$ after first equation?

@PK_editor it seems so - i've updated now
