Let consider two polynomials $f(x)$ and $g(x)$ with rational coefficients such that the degree of $f$ is divisible by the degree of $g$: $\deg f(x) = k \cdot \deg g(x)$, where $k$ is a natural number. What is the criterion for the polynomial $f$ to be a power of the polynomial $g(x)$, that is, when the equality $f(x) = (g(x))^k$ up to constant factor holds?
One approach is to find the resultants of all derivatives of $f$ and $g$: $$Res(f, g), Res(f', g), Res(f'', g), \ldots, Res(f^{(k-1)}, g).$$ If they are all equal to zero, then $f(x) = C (g(x))^k$, $C$ is a constant factor. But this is a long complicated method; is there a simpler way?
P.S. The polynomial $f$ is a linear combination of the known polynomials $f_1, f_2, \ldots, f_n$. I need to find such coefficients $\alpha_1, \alpha_2, \ldots, \alpha_n$ that $$f = \alpha_1 f_1 + \alpha_2 f_2 + \ldots + \alpha_n f_n = C \cdot (g(x))^k $$ Therefore, I cannot directly verify that $f(x) = C \cdot (g(x))^k$.
P.S.2 I have the folynomials $f_1, f_2, \ldots, f_n, g $ over $\mathbb{Q}$, $\deg f_1=\deg f_2 = \cdots = \deg f_n$. How can I decide if there are scalars $\alpha_i$ such that the linear combination of $f_i$ is a power of $g$ up to constant factor $$ f= \sum_{i=1}^n \alpha_i f_i=C \cdot (g(x))^{\frac{\deg f}{\deg g}}? $$