Prove that $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without Calculus. Prove that $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without Calculus. This inequality is equivalent to
$$
2018^{1/2018}>2019^{1/2019}
$$
One of my 'High school' student asked why the inequality is true. The whole class became interested in the problem.The demonstration that such inequality is true, using calculus, can be found here. But my students are not familiar with calculus.
I can also show by induction and Newton's binomial formula that $ n^{(n + 1)}> (n + 1)^n $, to $ n> 3$, but my students are not familiar with mathematical induction. 
Another limitation of my students is that they have not yet learned the  Newton's binomial formula.

How to prove the inequality $2018^{2019}> 2019^{2018}$ without induction, without Newton's binomial formula and without calculus? That is how to prove this inequality for High school students without using Newton's binomial formula?


Compared to my first answer, this answer makes a simpler use of the same familiar identity, which is proved without explicit use of induction, thus:
\begin{align*}
x^m - y^m & = x^m - (x^{m-1}y - x^{m-1}y) - (x^{m-2}y^2 - x^{m-2}y^2) - \cdots - (xy^{m-1} - xy^{m-1}) - y^m \\
& = (x^m - x^{m-1}y) + (x^{m-1}y - x^{m-2}y^2) + (x^{m-2}y^2 - x^{m-3}y^3) + \cdots + (xy^{m-1} - y^m) \\
& = (x - y)x^{m-1} + (x - y)x^{m-2}y + (x - y)x^{m-3}y^2 + \cdots + (x - y)y^{m-1} \\
& = (x - y)(x^{m-1} + x^{m-2}y + x^{m-3}y^2 + \cdots + y^{m-1}).
\end{align*}
The rest of the proof is a straightforward calculation. (It is based on a trick which was inspired loosely by nbarto's answer, but I am to blame for it!)
Suppose $n \geqslant 3$.
In the above identity, take $x = m = n+1$, $y = n$, and group together all but the first two terms in the brackets, obtaining:
$$
(n+1)^{n+1} - n^{n+1} = (n+1)^n + (n+1)^{n-1}n + [(n+1)^{n-2}n^2 + \cdots + n^n].
$$
There are $n-1$ terms in the square brackets, they are in strictly decreasing order, and the largest of them is $(n+1)^{n-2}n^2$. Therefore:
\begin{align*}
(n+1)^{n+1} - n^{n+1} & < (n+1)^n + (n+1)^{n-1}n + (n-1)(n+1)^{n-2}n^2 \\
& = (n+1)^n + (n+1)^{n-2}[n(n+1) + n^2(n-1)] \\
& = (n+1)^n + (n^3+n)(n+1)^{n-2}.
\end{align*}
On the other hand:
\begin{align*}
(n+1)^{n+1} - (n+1)^n & = n(n+1)^n \\
& = (n+1)^n + (n-1)(n+1)^n \\
& = (n+1)^n + (n-1)(n+1)^2(n+1)^{n-2} \\
& = (n+1)^n + (n^2-1)(n+1)(n+1)^{n-2} \\
& = (n+1)^n + (n^3 + n^2 - n - 1)(n+1)^{n-2}.
\end{align*}
But:
$$
(n^3 + n^2 - n - 1) - (n^3 + n) = n^2 - 2n - 1 = (n - 1)^2 - 2 > 0,
$$
therefore:
$$
(n+1)^{n+1} - n^{n+1} < (n+1)^{n+1} - (n+1)^n,
$$
therefore $n^{n+1} > (n+1)^n$. $\square$
