Suppose X follows the Pareto Type I given,
$$P(X>x) = \Bigl(\frac{\gamma}{x}\Bigl)^\alpha, \quad x\geq \gamma,\; \alpha>0.$$
Then, \begin{align*} P(X=x) &= \frac{d}{dx}P(X\leq x)\\ &= \frac{d}{dx}\biggl(1-\Bigl(\frac{\gamma}{x}\Bigl)^{\alpha}\biggl)\\ &= \frac{\alpha\gamma^\alpha}{x^{\alpha+1}}, \quad x\geq \gamma,\; \alpha>0. \end{align*}
Using the property of the order statistics, \begin{align*} f_{X_{(n)}}(x) &= nf(x)F(x)^{n-1}\\ &= n\cdot\frac{\alpha\gamma^\alpha}{x^{\alpha+1}}\biggl(1-\Bigl(\frac{\gamma}{x}\Bigl)^{\alpha}\biggl)^{n-1}, \quad x \geq \gamma,\; \alpha>0. \end{align*}
I'm trying to derive the expectation of the maximum order statistic, which is \begin{align*} E(X_{(n)}) &= \int_{\gamma}^\infty x\cdot n\cdot\frac{\alpha\gamma^\alpha}{x^{\alpha+1}}\biggl(1-\Bigl(\frac{\gamma}{x}\Bigl)^{\alpha}\biggl)^{n-1} dx\\ &= n\alpha \int_{\gamma}^{\infty} \Bigl(\frac{\gamma}{x}\Bigl)^{\alpha}\biggl(1-\Bigl(\frac{\gamma}{x}\Bigl)^{\alpha}\biggl)^{n-1} dx \end{align*}
but do not know how to proceed from this step. (also not sure is there any closed-form value for this expectation)
$$I=\int_{\gamma}^{\infty} \Bigl(\frac{\gamma}{x}\Bigl)^{\alpha}\biggl(1-\Bigl(\frac{\gamma}{x}\Bigl)^{\alpha}\biggl)^{n-1} dx$$
The antiderivative is given in terms of the Gaussian hypergeometric function $$-\frac{x }{\alpha -1}\left(\frac{\gamma }{x}\right)^{\alpha } \, _2F_1\left(1-n,\frac{\alpha -1}{\alpha };\frac{2\alpha -1}{\alpha };\left(\frac{\gamma }{x}\right)^{\alpha }\right)$$ Using the limits $$\color{blue}{I=\frac{\gamma }{\alpha }\,\,\frac{\Gamma \left(1-\frac{1}{\alpha }\right) \Gamma (n)} {\Gamma \left(n+1-\frac{1}{\alpha }\right) }}$$
Edit
Letting $x=\gamma \, t^{-1/a}$, you also have $$I=\frac \gamma \alpha \int_0^1 t^{-1/a} (1-t)^{n-1}\,dt$$ The antiderivative is $$\frac \gamma {\alpha-1} \,t^{\frac{\alpha -1}{\alpha }}\, _2F_1\left(1-n,\frac{\alpha -1}{\alpha };\frac{2\alpha-1}{\alpha};t\right)$$