One easy and insightful way is to use the proof below. It essentially constructs $\rm\:gcd\:$ from $\rm\:lcm\:$ by employing duality between minimal and maximal elements - see the Remark below. This is essentially how the linked Wikipedia proof works, but there the innate duality is obfuscated by the presentation. Below is a proof structured so that this fundamental duality is brought to the fore.
$\rm{\bf Theorem}\quad c\mid a,b\iff c\mid d,\ \ $ for $\rm\ \ d = ab/lcm(a,b).\ $ $\rm\color{#0a0}{Hence}$ $\rm\ d = gcd(a,b)$
$\rm{\bf Proof}\qquad\ \ \, c\mid a,b \iff a,b\mid ab/c \iff lcm(a,b)\mid ab/c \iff c\mid ab/lcm(a,b)$
$\rm\color{#0a0}{Generally}\,$ if $\rm\, c\mid a,b\iff c\mid d\ $ then $\rm\ d = \gcd(a,b)\ $ up to unit factors, i.e. they're associate.
Indeed setting $\rm\:c = d\:$ in direction $(\Leftarrow)$ shows that $\rm\:d\mid a,b,\:$ i.e. $\rm\:d\:$ is a common divisor of $\rm\:a,b.\:$ Conversely, by direction $(\Rightarrow)$ we deduce that $\rm\:d\:$ is divisible by every common divisor $\rm\:c\:$ of $\rm\:a,b,\:$ thus $\rm\:c\mid d\:\Rightarrow\: c\le d,\:$ so $\rm\:d\:$ is a greatest common divisor (both divisibility and magnitude-wise).
Remark $\ $ The proof shows that, in any domain, if $\rm\:lcm(a,b)\:$ exists then $\rm\:gcd(a,b)\:$ exists and $\rm\ gcd(a,b)\,lcm(a,b) = ab\ $ up to unit factors, i.e. they are associate. The innate duality in the proof is clarified by employing the involution $\rm\ x'\! = ab/x\ $ on the divisors of $\rm\:ab.\:$ Let's rewrite the proof using this involution (reflection).
Notice that $\rm\ x\,\mid\, y\:\color{#c00}\iff\: y'\mid x'\,\ $ by $\smash[t]{\,\ \rm\dfrac{y}x = \dfrac{x'}{y'} \ }$ by $\rm\, \ yy' = ab = xx',\ $ so rewriting using this
$\begin{eqnarray}\rm the\ proof\ \ \ c\mid a,b &\iff&\rm b,\,a\mid ab/c &\iff&\rm lcm(b,\,a)\mid ab/c &\iff&\rm c\mid ab/lcm(b,a)\\[.5em]
\rm becomes\ \ \ \ c\mid a,b &\color{#c00}\iff&\rm a',b'\mid c' &\iff&\rm lcm(a',b')\mid c' &\color{#c00}\iff&\rm c\mid lcm(a',b')'\end{eqnarray}$
Now the innate duality is clear: $\rm\ gcd(a,b)\,=\,lcm(a',b')'\ $ by the $\rm\color{#0a0}{above}$ gcd characterization.
Expanding $\,(n,a\!-\!b)(n,a\!+\!b) = n\,(\overbrace{\color{#c00}{n,a\!-\!b,a\!+\!b},(a^2\!-\!b^2)/n}^{\large =\ 1\ \rm by\ below}) = n$
since $\,p\mid \color{#c00}{n,a\!-\!b,a\!+\!b}\,\Rightarrow\, p\mid n,2a,2b,\,$ contra $\,n\,$ coprime to $\,2,a,b\,$ (wlog)
Remark $ $ The factorization $\,n = (n,a\!-\!b)(n,a\!+\!b)\,$ is proper when $\,n\nmid a\pm b,\,$ so then $\!\bmod n\!:\ b^2\equiv a^2\,$ but $\,b\not\equiv \pm a,\,$ i.e. $\,b\,$ is a nontrivial square root of $\,a^2,\,$ so the quadratic $\,x^2-a^2$, has more than $\:\!2\:\!$ roots (assuming wlog $\,n\,$ is odd).
Generally we can quickly split $\,n>1\,$ into two nontrivial factors given any nonzero polynomial with more roots $\!\bmod n\,$ than its degree - see here.
Best Answer
(1) Next, consider any common divisor $c$, and make a note about how the exponents in its prime factorization compare to min$(\alpha_i, \beta_i)$.
(2) From here, consider what would happen if the greatest common divisor could not be written in the form $c^nr$. You can arrive at a contradiction by using the definition of gcd.