Proving "$\forall_{\epsilon>0} \exists_{N \in \mathbb{N}} \forall_{n \geq N}: x^{n} < \epsilon$" for $0I'm trying to prove a rather simple analysis statement but I think I'm overseeing something. For $0<x<1$ I need to prove

$$\forall_{\epsilon>0} \exists_{N \in \mathbb{N}} \forall_{n \geq N}: x^{n} < \epsilon.$$

I now have the following:
We use theorem 1.5 which states 
\begin{equation}\label{theorem_1.5}
    \forall_{\epsilon>0}, \exists_{N \in \mathbb{N}} : \frac{1}{N} < \epsilon.
\end{equation}
Also we make use of the fact that if $0<x<1$ it follows that 
\begin{equation}\label{equation_1}
    (N+1)x^{N}<\frac{1}{1-x}.
\end{equation}
Suppose $0<x<1$ and let $\epsilon>0$ be given. Now let $\epsilon'=\frac{1}{1-x}\epsilon$. Then from equation 1, equation 2 and the fact that $N+1>N$ it follows that 
\begin{equation}
    \begin{split}
    (N+1)x^{N} &< \frac{1}{1-x},\\
    x^{N} &< \frac{1}{1-x}\frac{1}{N+1} < \frac{1}{1-x}\frac{1}{N}\\
    x^{n} &< \frac{1}{1-x} \epsilon\\
    x^{n} &< \epsilon'.
    \end{split}
\end{equation}
My gut feeling says there is something wrong with my reasoning. It might be the fact that I predetermine $\epsilon'$, but I'm not sure.
Any thoughts would be appreciated.

Put $y=1/x$, then $y>1$, hence $y=1+z$ for some $z>0$. Then we get, with Bernoulli:
$$\frac{1}{x^n}=y^n=(1+z)^n \ge 1+nz >nz.$$
Hence
$$ x^n < \frac{1}{nz}.$$
Can you proceed ?
