Prove that $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without Calculus.

I would try to motivate this by showing that $f(x) = x^{1/x}$ is a monotone function for reasonably small $x$.

Another approach is to note that $$ \frac{2019^{2018}}{2018^{2018}} = \left(1 + \frac{1}{2018}\right)^{2018} $$ and so your inequality is equivalent to showing $$ \left(1 + \frac{1}{2018}\right)^{2018} < 2018, $$ which does not sound very far-fetched, since LHS is close to $e$...

UPDATE

Please see saulspatz's answer for how to prove this last claim with a hand computation only.

$$1+{1\over2018}<1+{1\over2000}=1.0005$$ $$\left(1+{1\over2018}\right)^{2018}<(1.0005)^{2048}$$

The right hand side can be evaluated by repeated squaring, and, since there is plenty of leeway, you can simplify the calculations by rounding up as you go. Since $(1.0005)^{2048}<3,$ this calculation should be well within you students' abilities.

Note that $2019^{2048}<2018^{2049}$ implies that $$2019^{2018}2019^{30}=2019^{2048}<2018^{2049}<2018^{2019}2019^{30},$$ which implies that $2019^{2018}<2018^{2019}$. We look at $2048$ in the exponent because it is a power of $2$.

Claim: $2019^{2048}<2018^{2049}$.

Proof: $$2019^{2048}-2018^{2048}=$$ $$(2019^{1024}+2018^{1024})(2019^{512}+2018^{512})...(2019^{2}+2018^{2})(2019+2018)(2019-2018)$$ by repeatedly factoring differences of squares. Each term of the form $2019^i+2018^i<2 \cdot 2019^i$, so taking each of these inequalities into account, we get that $$2019^{2048}-2018^{2048}<2^{10}\cdot 2019^{2047}$$ since $1+2+4+8+...+512+1024=2^{11}-1$. Then we combine terms with like bases to get that $$2019^{2048}-1024\cdot 2019^{2047}=995\cdot 2019^{2047}<2018^{2048}.$$ We now multiply both sides by $2018$ to get $$2019^{2048}<995\cdot 2018\cdot 2019^{2047}<2018^{2049},$$ which is the desired result.