If $p>1$, then I do not believe the LHS will converge over the reals. Perhaps you are supposed to solve for $p$, and it is not greater than 1? In other words, you are looking for:
$$\int_0^1 \dfrac{dx}{x^p} = 1$$
This integral is fairly well-known. If $p\ge 1$, then you have the integral does not converge. If $p\neq 0$, then you have:
$$\dfrac{1^{1-p}}{1-p}-\dfrac{0^{1-p}}{1-p}=1 \Longrightarrow p=0$$
So, that means that $p=0$ is the only solution.
Not only does it help, but it is necessary. Otherwise the statement would be false.
Do you understand the theorem I quoted.
We have $\displaystyle 0 < 1 < 2 < \cdots < n$ so by the theorem
$\displaystyle \int_0^n {\frac{{dx}}{{{x^p}}}} = \int_0^1 {\frac{{dx}}{{{x^p}}}} + \int_1^2 {\frac{{dx}}{{{x^p}}}} + \cdots + \int_0^n {\frac{{dx}}{{{x^p}}}} = \sum\limits_{k = 1}^n {\int_{k - 1}^k {\frac{{dx}}{{{x^p}}}} } $
Now $\displaystyle \int_0^1 {\frac{{dx}}{{{x^p}}}} = \mathop {\left. {\frac{{{x^{1 - p}}}}{{p - 1}}} \right|}\nolimits_0^1 =1$
I agree, how is that known? I tried it for $p=2$ and according to Wolframalpha, the integral did not converge.
Wolframalpha