Prove in as elementary a way as possible that $a^{1/(a-1)} < 2$ for integer $a \ge 3$ This inequality came up
as the final step
in an elementary proof
that the only 
integer solution to
$x^y = y^x$
with $x > y > 1$
is
$4^2 = 2^4$.
Here are two proofs
I came up with that
if integer $a \ge 3$,
then $ a^{1/(a-1)} < 2$.
Both proofs do it
by noting that
this is true for
$a=3$,
and $a^{1/(a-1)}$
is decreasing for
$a \ge 3$.
One way to show this
is by looking at
the log of this,
which is
$\frac{\ln a}{a-1}$.
Its derivative is
$\frac{(a-1)/a - \ln a}{(a-1)^2}
=\frac{1-1/a - \ln a}{(a-1)^2}
< 0$
for $a \gt 1$.
If 
$f(a) =1-1/a - \ln a
$
then
$f(1) = 0$
and
$f'(a)
=1/a^2-1/a < 0
$ for $a > 1$.
Another proof:
Suppose 
$a^{1/(a-1)} \le (a+1)^{1/a}
$.
Then
$a^a \le (a+1)^{a-1}$
or
$(a+1)a^a \le (a+1)^a$.
Dividing by
$a^a$,
$a+1 \le (1+1/a)^a$.
But,
as has been proved here
by elementary means
many times,
$(1+1/a)^a < 3$,
so that
$a+1 < 3$
or
$a < 2$.
So,
is there a more elementary proof than these?
Thanks.

The inequality $a^{1/(a-1)}<2$ for integers $a\ge 3$ is equivalent to $a<2^{a-1}$ for integers $a\ge 3$. This is easily proved by induction on $a$. It’s clearly true for $a=3$, and if $a<2^{a-1}$ and $a\ge 3$, then
$$a+1<2^{a-1}+1<2\cdot 2^{a-1}=2^a\;.$$
