Is $(\left(\frac{1}{x}\right)^{\frac{1}{p} - \frac{1}{n}})_{n\in N}$ a Cauchy Sequence in $L^p((0,1))$? and does it converge to $\frac{1}{x}^{\frac{1}{p}}$ (p is a real number bigger or equal to 1)
I think that it isn't a cauchy sequence or doesn't converge to $\frac{1}{x}^{\frac{1}{p}}$, because if it was, then $\frac{1}{x}^{\frac{1}{p}}$ would be in $L_p((0,1))$, because the $L^p$ spacess are complete and
$$\left(\int_{(0,1)} \left|\left(\frac{1}{x}\right)^{\frac{1}{p} - \frac{1}{n}}\right|^p dL_1(x)\right)^{1/p} = \left(\int_{(0,1)} \left|\left(\frac{1}{x}\right)^{1 - \frac{p}{n}}\right| dL_1(x)\right)^{1/p} < \infty$$
but
$$\int_{(0,1)} \left(\left(\dfrac{1}{x}\right)^{\frac{1}{p}}\right)^p dL_1(x) = \int_{(0,1)} {\frac{1}{x}} dL_1(x)$$
which diverges
As you observe, the limit, if it exists, cannot be $(\frac{1}{x})^{\frac1p}$. That is the pointwise limit, though. If that sequence had a limit in $L^p$, we would have a subsequence converging pointwise almost everywhere to that limit, which couldn't be the actual pointwise limit, and the real pointwise limit would therefore not be the pointwise limit. Let me explain this better. Call the sequence $f_n$. Suppose $f_n\to f$ in $L^p$. Then we know that there exists $f_{n_k}$ a subsequence of $f_n$ that converges pointwise almost everywhere to $f$. $f$, however, could not be $g(x)=(\frac1x^{\frac1p})$, since $g\not\in L^p$, as you showed. So we would have a subsequence of $f_n$ pointwise converging almost everywhere to something different from the real pointwise limit $g$, which could therefore not be the pointwise limit of the whole sequence. This is an evident contradiction. Therefore, $f_n$ can't converge in $L^p$. Since the $L^p$ spaces are complete, it cannot be a Cauchy sequence either, otherwise it would converge, etc etc etc.
A more "constructive" or practical disproof of this statemente (that $f_n$ converges or is Cauchy in $L^p$) is to show it is not bounded. Let's take the $L^p$ norm of $f_n$: \begin{align*} \|f_n\|_p^p={}&\int\limits_{[0,1]}\left|\left(\frac1x\right)^{\frac1p-\frac1n}\right|^p\mathrm{d}x=\int\limits_{[0,1]}\left(\frac1x\right)^{1-\frac pn}\mathrm{d}x=-\int\limits_{[+\infty,1]}t^{1-\frac pn}\frac{1}{t^2}\mathrm{d}t={} \\ {}={}&-\frac{1}{-\frac pn}t^{-\frac pn}\Bigg|_{+\infty}^1=-\frac{1}{\frac pn}\lim_{t\to+\infty}t^{-\frac pn}+\frac{1}{\frac pn}=\frac np, \end{align*} which for $n\to\infty$ is unbounded, thus proving the sequence cannot converge. Indeed, if it did converge to a limit $f\in L^p$, we would have $\|f_n-f\|_p\to0$, but $\|f_n-f\|_p\geq\|f_n\|_p-\|f\|_p$ which is unbounded for $n\to\infty$.