Euclid first proved the infinitude of primes. For those who don't know, here's his proof:
Let $p_1=2,p_2=3,p_3=5,...$ be the primes in ascending order, and suppose there is a last prime, call it $p_n$. Now consider the positive integer $$P=p_1p_2p_3....p_n+1$$ Because $P>1$, the Fundamental theorem of Arithmetic tells us that $P$ is divisible by some prime $p$. But $p_1,p_2,...p_n$ are the only prime numbers, so $p$ must be equal to one of $p_1,p_2,...p_n$. Combining the divisibility relations $p|p_1p_2...p_n$ with $p|P$, we get $p|P-p_1p_2...p_n$ or equivalently, $p|1$, so $p=1$ or $p=-1$, which is absurd. So $p$ is not one of $p_1,p_2,...p_n$, so the list of primes never ends.
Many people (like me) are not content with only one proof. An alternative proof was given by Euler:
For a prime $p$, the ratio $\frac{1}{(1−1/p)}$ can be expanded into a geometric series: $$\frac{1}{(1−1/p)}=1+\frac{1}{p}+\frac{1}{p^2}+\frac{1}{p^3}+...$$ We will now multiply together these ratios for different primes. To see what can be obtained, let’s look at the product of these terms for the primes $2$, $3$, and $5$: $\frac{1}{1−1/2}\frac{1}{1-1/3}\frac{1}{1-1/5}$ equals $$\left ( 1+\frac{1}{2}+\frac{1}{4}+... \right )\left (1+\frac{1}{3}+\frac{1}{9}+... \right )\left ( 1+\frac{1}{5}+\frac{1}{25}+... \right )$$ and if we multiply together one term from each series (this is the distributive law for multiplying infinite series, and it is valid when multiplying convergent series of positive numbers) we will get unit fractions $1/n$ where $n$ is a product of powers of $2, 3$, and $5$: $$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{12}+\frac{1}{15}+\frac{1}{16}+...$$ This looks like the harmonic series, but it is missing terms 1/n where n has a prime factor greater than $5$, such as terms $1/7, 1/11$, and $1/14$. If we multiply by additional factors $1/(1 − 1/p)$ for more primes $p$ we’ll introduce into the harmonic-like series new terms $1/n$ where $n$ has prime factors involving the new primes and the ones we already used ($2, 3$, and $5$). Because of unique factorization in $\mathbb{Z}^+$, each term $1/n$ that appears will do so just once. Therefore if we multiply together $1/(1 − 1/p)$ for all the primes, the harmonic-like series turns into the harmonic series: $$\prod_{p\,\text{prime}}\frac{1}{1-p^{-1}}=\sum_{k\geq1}\frac{1}{k}$$ Since the RHS diverges, the LHS also diverges, but since there are only finitely many primes, the LHS has to be convergent, which is a contradiction. So, there are infinitely many primes.
But, I would like to see more proofs. So this question is to give as many proofs as you can. So please give some more proofs. Every answer will be appreciated.