Prove that $\Sigma a_\nu = \Sigma b_\nu$ (Remmert, exercise 2, section 1 chapter 0)

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I'm studying Complex analysis from Remmert Theory of Complex Functions book and I'm trying to do the exercises. However, I'm stuck with exercise 2, section 1, chapter 0. It says:

Let $a_1,...,a_n, b_1,...b_n\in \mathbb{C}$ and satisfy $\sum_{\nu=1}^{n} a^j_\nu=\sum_{\nu=1}^{n} b^j_\nu$ for all $j \in \mathbb{N}$. Show that there is a permutation $\pi$ of $\{1,2,...n\}$ such that $a_\nu=b_{\pi(\nu)}$ for all $\nu \in \{1,..., n\}.$

I tried to see $\sum_{\nu=1}^{n} a^j_\nu=\sum_{\nu=1}^{n} b^j_\nu$ as a polynomial of $2n$ variables like $p(x_1,...,x_n,y_1,...,y_n)= \sum_{\nu=1}^{n} x^j_\nu - y^j_\nu$ and express it like product of its roots. However, I need to use the Fundamental Theorem of Symmetric Polynomials, and I do not know where to use it!.

I'm so confused in this exercise. I will apreciate all your help!