Iterated exponentials in modular arithmetic: $a^{b^{c^d}}-a^{b^c}\equiv 0 \mod 2^{2^{2^2}}-2^{2^2}$? How to prove $a^{b^{c^d}}-a^{b^c}\equiv 0 \mod 2^{2^{2^2}}-2^{2^2}$ for all positive integers $a,b,c,d\in \mathbb{N}$?  I tried to reduce $d$ to $2$ by induction. But I kind of stuck at the rest of the parameters. Any thoughts?

Clearly, this is true if at least one of $b,c,d$ is $1$, so we can safely focus 
on $b,c,d>1$.
$2^{2^{2^2}}-2^{2^2}=2^{16}-2^4=2^4(2^{12}-1)=2^4(2^6-1)(2^6+1)=2^4\cdot7\cdot 9\cdot 13\cdot 5$
We know,for any prime $p$ either $p\mid a$ or $(p,a)=1$
So, if $p\mid a,p\mid(a^e-a)$ for $e\ge 1$  else  $(p,a)=1,$
(i)For $p=13,\phi(13)=12$
if $13\mid a,13\mid(a^{b^{c^d}}-a^{b^c})$
if $(13,a)=1$ it is sufficient to  prove $12\mid(b^{c^d}-b^c)$
Now, $b^{c^d}-b^c=b^c(b^{c^d-c}-1)=b^c(b^{c(c^{d-1}-1)}-1)$
As $c,d>1$ so, $b^2(b^2-1)\mid b^c(b^{c(c^{d-1}-1)}-1)$
Now, if $2\mid b,2^2\mid b^2$ else $b$ is odd$=2c+1$(say),$b^2-1=(2c+1)^2-1=8\frac{c(c+1)}2$  
and $3\mid (b-1)b(b+1)$.
So, $12\mid b^2(b^2-1)\implies 12\mid  b^c(b^{c(c^{d-1}-1)}-1)$
(ii) For $p=7,\phi(7)=6$  
if $7\mid a, 7\mid(a^{b^{c^d}}-a^{b^c})$
if $(7,a)=1,$ it is sufficient to  prove $6\mid(b^{c^d}-b^c)$
(iii) For $p=5,\phi(5)=4$ 
if $5\mid a, 5\mid(a^{b^{c^d}}-a^{b^c})$
if $(5,a)=1,$ it is sufficient to  prove $4\mid(b^{c^d}-b^c)$
$2^4$ and $9=3^2$  need different approach as they are powers$(>1)$  of primes.
(iv)For $2^4,$
If $2\mid a,$ both $b^c$ and $b^{c^d}\ge 4$ as $b,c,d>1$,
hence, both $a^{b^c}$ and $a^{b^{c^d}}$ are divisible by $2^4$
If $(2,a)=1,\lambda(16)=\frac{\phi(16)}2=4,$ it is sufficient to  prove $4\mid(b^{c^d}-b^c)$ where $\lambda$ is the Carmichael Function.
(v)For $9,$ 
If $3\mid a,$ both $b^c$ and $b^{c^d}\ge 4$ as $b,c,d>1$,
hence, both $a^{b^c}$ and $a^{b^{c^d}}$ are divisible by $3^4,$ hence by $3^2=9$
If $(3,a)=1,\phi(9)=6,$ it is sufficient to  prove $6\mid(b^{c^d}-b^c)$
So, in all the cases, $a^{b^{c^d}}-a^{b^c}$ is divisible by $2^4, 7, 9, 13, 5;$ hence by $lcm(2^4, 7, 9, 13, 5)=2^4\cdot7\cdot 9\cdot 13\cdot 5=2^{2^{2^2}}-2^{2^2}$ 
