How do i prove there exists a function $F(n+1)=f_n(F(1),...,F(n))$? Let $X$ be a set and $f_n:X^n \rightarrow X$ be a function for all $n\in \mathbb{N}$ and $c\in X$.

How do i prove that there exists $F:\mathbb{N} \rightarrow X$ such that $F(n+1)=f_n(F(1),...,F(n))$ and $F(1)=c$?

Plus, is this $F$ unique?
The quick brute-force method would be to use Zorn's lemma.

Consider the set of all functions $A\to X$ where $A$ is an initial sequence of $\mathbb N$ and the function satisfies your equation. Order the functions by set inclusion. Zorn's lemma says there is a maximal element. The domain of this maximal element cannot have a largest element, because then you would easily construct a function that went one higher. Therefore the domain must be $\mathbb N$ itself, and the maximal element is your $F$.

(It can be done without Zorn or choice, but there will be more thinking involved then).

For uniqueness, suppose there are $F$ and $G$ that both satisfy your equation. Prove by (long) induction on $n$ that $F(n)=G(n)$ for all $n$, and therefore $F=G$.
The quick brute-force method would be to use Zorn's lemma.

Consider the set of all functions $A\to X$ where $A$ is an initial sequence of $\mathbb N$ and the function satisfies your equation. Order the functions by set inclusion. Zorn's lemma says there is a maximal element. The domain of this maximal element cannot have a largest element, because then you would easily construct a function that went one higher. Therefore the domain must be $\mathbb N$ itself, and the maximal element is your $F$.

(It can be done without Zorn or choice, but there will be more thinking involved then).

For uniqueness, suppose there are $F$ and $G$ that both satisfy your equation. Prove by (long) induction on $n$ that $F(n)=G(n)$ for all $n$, and therefore $F=G$.