How existence of an irreducible polynomial in $F_p(x)$ of degree $n$ guarantees existence of finite field of order $p^n$ I read somewhere that if $\pi$ is an irreducible polynomial of degree $m$ then $F_p(x)\ \backslash \left< \pi \right>$ is a finite field of order $p$. What is $F_p(x)\ \backslash \left< \pi \right>$? How does it guarantee existence of a finite field of order $p^m$?
Here is my understanding of the answer:
Say $\pi$ is an irreducible polynomial of degree $m$. Consider any two polynomials $p_1$ and $p_2$ each of degree less than $m$. Say $p_1(x) = a_0 + a_1x +...+a_{m-1}x^m$ and $p_2(x) = b_0 + b_1x + ...+ b_{m-1}x^{m-1}$. Since polynomials uniquely factor into irreducible polynomials $\pi$ can't divide the product of these two polynomials since the factorization of the product contains irreducible polynomials of degree at most $m-1$. Therefore $(p_1\times p_2)/\pi(x)$ is a polynomial of degree less than $m$. 
Therefore for polynomials $p_1, p_2$ of degree less than $m$, we define $p_1\cdot p_2 = (p_1\times p_2)/\pi$ and for non-zero polynomials it is guaranteed that this $p_1\cdot p_2$ is not equal to zero. 
This also means that for every non-zero polynomial $p_1$ of degree less than $m$, $p_1\cdot p_2$ is not equal to $p_1 \cdot p_3$ if $p_2$ is not the same as $p_3$. Otherwise $(p_1\times p_2)/\pi-(p_1\times p_3)/\pi=((p_1\times p_2)-(p_1\times p_3))/\pi=(p_1\times (p_2-p_3))/\pi = 0$ implying that $p_2 = p_3$, a contradiction. 
This also guarantees existence of multiplicative inverse since for exactly one polynomial $u$ we must have $p_1\cdot u=1$.
Therefore if multiplication is defined this way, then all axioms of the field are satisfied and hence we've obtained a field.    

If $p$ is irreducible in $F[X]$ ($F$ being any field), the quotient ring $F[X]/(p)$ is a field because an irreducible polynomial generates a maximal ideal of $F[X]$, and an $F$-vector space of dimension $n=\deg p$.
If $F=\mathbf F_p$ is the prime field with characteristic $p$, a vector space of dimension $n$ has exactly $p^n$ elements, and this one is a field. If we denote $x=X+(p)$,this field is denoted $\mathbf F_p(x)$.
