OK, now that the standard theory's up, I'll add my favorite argument. We prove that irreducible polynomials of every degree exist by counting them.
Let $F$ be a field with $q$ elements, for some finite $q$. We count the number of monic polynomials of degree $n$ over $F$ in two ways.
First, a monic polynomial $x^n + a_{n-1}x^{n-1}+\cdots + a_1 x + a_0$ has $n$ coefficients that can freely vary among $q$ values each, for a total of $q^n$ polynomials.
Second, a monic polynomial can be uniquely factored as a product of monic irreducible polynomials. We choose the power each polynomial is raised to, make sure the sum of the degrees is $n$ - it's an intimidating counting problem, but we can make it easier by thinking in terms of generating functions. Let $P_k$ be the number of monic irreducible polynomials of degree $k$, and $M_n$ the total number of monic polynomials of degree $n$. This model gives us that
$$\sum_{n=0}^{\infty} M_n t^n = \prod_{k=1}^{\infty}\frac1{(1-t^k)^{P_k}}$$
(Each $\frac1{1-t^k}=1+t^k+t^{2k}+\cdots$ factor represents the choice of how many copies of a particular irreducible polynomial of degree $k$ to include)
So, what now? First, we use that earlier count to simplify the left side:
$$\sum_{n=0}^{\infty} M_n t^n = \sum_{n=0}^{\infty} q^n t^n = \frac1{1-qt}$$
Next, take logarithms* and differentiate with respect to $t$:
\begin{align*}\frac1{1-qt} &= \prod_{k=1}^{\infty}\frac1{(1-t^k)^{P_k}}\\
-\log(1-qt) &= \sum_{k=1}^{\infty} -P_k\log(1-t^k)\\
\frac{q}{1-qt} &= \sum_{k=1}^{\infty} \frac{kP_kt^{k-1}}{1-t^k}\\
\sum_{n=1}^{\infty} q^nt^n = \frac{qt}{1-qt} &= \sum_{k=1}^{\infty} \frac{kP_kt^k}{1-t^k} = \sum_{n=1}^{\infty}t^n\sum_{k|n}kP_k\\
q^n &= \sum_{d|n}dP_d\end{align*}
Now, we can extract the values of $P_n$ by a standard technique; Mobius inversion. We get
$$nP_n = \sum_{d|n} \mu(d)q^{\frac{n}{d}}$$
where $\mu$ is the Mobius function of number theory.
Now, what we really wanted to know is that $P_n$ is positive for $n\ge 1$. For this, one key fact: $\mu(d)\in \{-1,0,1\}$ for all natural $d$ and $\mu(1)=1$, so
\begin{align*}nP_n &= q^n +\sum_{k|n, 0<k<n}\mu(\frac{n}{k})q^k\\
nP_n &\ge q^n - \sum_{k|n, 0<k<n} q^k \ge q^n - \sum_{0<k<n}q^k = q^n - \frac{q^n-q}{q-1} > 0\end{align*}
We use that $q\ge 2$ here; we do need at least two elements to have a field, after all. And there it is: any finite field with $q$ elements has irreducible polynomials for every degree. This includes both the prime fields $\mathbb{Z}/p$ and fields with prime power order. That such fields exist to have irreducible polynomials over - well, we'll just have to apply this result over the prime fields first.
*Taking a logarithm? Technically, it's all still formal power series manipulation, that doesn't rely on any form of convergence - but if you feel uncomfortable, we're just taking a derivative and then dividing by the original.
