Justifying $\int_{0}^{1}\frac{\partial^n}{\partial m^n} x^m dx=\frac{\partial^n}{\partial m^n}\int_{0}^{1} x^m dx$

Question

$$\int_{0}^{1}x^m (\log(x))^n dx=\int_{0}^{1}\frac{\partial^n}{\partial m^n} x^m dx=\frac{\partial^n}{\partial m^n}\int_{0}^{1} x^m dx$$

How can I justify the last step I made there? Dominate convergence theorem?