Convergence in distribution for exponent sequence Let $X_1, X_2\ldots$ be random variables with $X_n \overset{d}{\rightarrow} X$ and $P(0 < X < 1) = 1$. 
Show that $X_n^{n} \overset{d}{\rightarrow} 0$.
I think if we use Skorohods Theorem we get $Y,Y_1, Y_2, \ldots$ random variables on the same probability space with $Y_i \overset{d}{=} X_i$, $Y \overset{d}{=} X$ and $Y_i \rightarrow Y$ P-a.s. That means the condition above is also true for $Y$ wich means $P(0<Y<1) = 1$. But now I'am stuck.
 A: Take $0<x<1$ and consider 
$$
F_{X_n^n}(x)=\mathbb P(X_n \leqslant \sqrt[n]{x}) = F_{X_n}(\sqrt[n]{x})
$$
Since $\sqrt[n]{x}\uparrow 1$ then for each $0<\varepsilon<1$ there exists $N=N(\varepsilon)$ s.t. for any $n\geqslant N$
$$
\sqrt[n]{x} \geqslant 1-\varepsilon.
$$
Then monotonicity of cdf implies that for $n\geqslant N$
$$
F_{X_n^n}(x)=F_{X_n}(\sqrt[n]{x}) \geqslant F_{X_n}(1-\varepsilon).
$$
If $F_X(x)$ is continuous in the point $1-\varepsilon$, then r.h.s. tends to $F_X(1-\varepsilon)$, and 
$$
1\geqslant\limsup_{n\to\infty}F_{X_n^n}(x)\geqslant\liminf_{n\to\infty}F_{X_n^n}(x) \geqslant F_X(1-\varepsilon).
$$
And the number of discontinuity points of every cdf is at most countable, so we can chose a sequence $\varepsilon_m\to 0$ s.t. $F_X(x)$ is continuous in all the points $1-\varepsilon_m$. Tends $m\to\infty$ and get 
$$
1\geqslant\limsup_{n\to\infty}F_{X_n^n}(x)\geqslant\liminf_{n\to\infty}F_{X_n^n}(x) \geqslant F_X(1-\varepsilon_m)\to F_X(1-0)=\mathbb P(X<1)=1.
$$
We prove that $F_{X_n^n}(x)\to 1$ for any $0<x<1$. And it is obvious that $F_{X_n^n}(x)\to 0$ for $x\leq 0$. So $X_n^n$ converges to zero in distribution.
The other way was to use Slutsky's theorem: for $0<x<1$, $\sqrt[n]{x}\to 1$, so 
$X_n-\sqrt[n]{x}\xrightarrow{d} X-1$ and 
$$
\mathbb P(X_n-\sqrt[n]{x}\leq 0) \to \mathbb P(X-1\leq 0)=F_X(1)=1.
$$
Since $F_X$ is continuous at $1$, $F_{X-1}$ is continuous at $0$ and convergence takes place. 
