How to prove $\log n < n$?
If we let $f(n) = \log n$ and $g(n) = n$, then $f'(n) = \frac{1}{n}$ and $g'(n)=1$. Since $f(1) < g(1)$ and $f'(n) \leq g'(n)$ for $n \geq 1$, we must have $f(n) < g(n)$ for all $n \geq 1$.
The idea in the last sentence is sometimes called the racetrack principle: If $f(n)$ and $g(n)$ denote the positions of horses $f$ and $g$ at time $n$, horse $f$ starts behind horse $g$ (i.e, $f(1) < g(1)$), and at any given time horse $f$ is never faster than horse $g$ (i.e, $f'(n) \leq g'(n)$) then horse $f$ will always be behind horse $g$ (i.e, $f(n) < g(n)$ for $n \geq 1$).
OP asks for a proof of the racetrack principle.
The racetrack principle: If $f(a) < g(a)$ and $f'(n) \leq g'(n)$ for $n \geq a$, then $f(n) < g(n)$ for $n \geq a$.
Proof: Let $h(n) = f(n) - g(n)$. Then $h'(n) = f'(n) - g'(n) \leq 0$. The mean value theorem tells us that there exists some point $x \in [a,n]$ such that $$h'(x) = \frac{h(n) - h(a)}{n-a}.$$ Since we know the claim is true for $n=a$, take $n > a$. We already know $h'(x) \leq 0$, so we get $h(n) - h(a) \leq 0$, which means $f(n) - g(n) \leq f(a) - g(a) < 0$, and so $f(n) < g(n)$ for $n \geq 1$.
Just simply look at the function $f(t)=t-\log(t)$. You can show that this function is always increasing and that $f(n)\ge f(1)=1$ for every $n$.
How rigorous do you need? If it's rigorous enough in your context, prove that $\log(1) < 1$ (shouldn't be hard!), and then show that the derivative of $\log(x)$ is less than or equal to the derivative of $x$ for all $x \geq 1$.