Sum of the digits in base $p+1$ 

Definition
Let $W$ be the function ,  defined as $W(a,b)=r$ 
given $a,b\in \mathbb{Z_+}$ and $a>1$
Take $m$ to be the integer s.t. $a^{m+1} \ge b > a^{m}$, i.e. $m = \lceil \log{b}/\log{a} \rceil - 1$.
Convert number $a^{m+1} - b$ in base $a$ and add its digits
$$a^{m+1} - b = (r_{l} r_{l-1} ... r_{1} r_{0})_{a}$$
Where $r=\sum_{i=0}^{l}r_{i}$


Example
$W(5,77)=8$
Identity$1$
if $W(a,b)=r$ then $b+r\equiv 1($ mod $a-1)$
◆ $S$ is a function defined as
$$S(a,n)=\sum_{i=1}^{a}i^{n}$$
Where $a$ and $n$ are positive integer.
Let $p$ is prime
and $p+1=z$

Question

show that
If $ z>2n+2$ Then $W(z,W(z,S(z,2n)))=z$


Example
Let $n=1$ here, choose any $z>4$
Let $z=6$ 
So $W(6,W(6,S(6,2)))=W(6,W(6,91))=W(6,10)=6$
Table For $W(t,W(t,S(t,2)))$.
$$\begin{array}{c | c | c |c | }  t & W(t,S(t,2)) & W(t,W(t,S(t,2))) \\ \hline
2 & 2  & 0 \\ \hline
3^{*} & 3 & 0 \\    \hline 
4^{*} & 4 & 0 \\ \hline
5 & 6 & 7 \\ \hline
6^{*} & 10 & 6 \\ \hline
7 &5 & 2 \\ \hline
8^{*} &14& 8 \\ \hline
9 &12& 13 \\ \hline
10 &12& 16 \\ \hline
11 & 15  & 16 \\ \hline
12^{*} & 22 & 12 \\    \hline 
13 & 10 & 3 \\ \hline
14^{*} & 26 & 14 \\ \hline
15 & 21 & 22 \\ \hline
16 &20 & 26 \\ \hline
17 &24& 25 \\ \hline
18^{*} &34& 18 \\ \hline
19 &15& 4 \\ \hline
20^{*} &38& 20 \\ \hline
21 &30& 31 \\ \hline
\vdots &\vdots & \vdots \\ \hline
 \end{array}$$ 
$t^{*} = z $
Python programming for calculate $W$ function
n1=5
n2=77
rem_array = []
while n2 != 1:
    mod = n2%n1
    if mod != 0:
      rem = n1-mod
      n2 = n2 + rem
      rem_array.append(round(rem))
      n2=n2/n1
    else:
        n2 = n2/n1
        rem_array.append(0)
print(rem_array[::-1])
print(sum(rem_array))



Proof for, if $p>n+1$ then $p|S(p,n)$ 

Formula
$$ S(a,n)= \sum_{i=1}^{a} i^{n}=\sum_{b=1}^{n+1} \binom{a}b\sum_{j=0}^{b-1} (-1)^{j}(b-j)^{n}\binom{b-1}j$$
for formula
Proof
Let $a=p(prime)>n+1$
We can see, $a$ can be common out from $\sum_{b=1}^{n+1}\binom{a}b\sum_{j=0}^{b-1} ...$
$\implies a|S(a,n)$

Proof for, If $ p|S(p,2n)$ Then $W(z,W(z,S(z,2n)))=(z-1)r+1=pr+1$ 

Proof
See $S(z,2n)=pr_1+1$
$\implies W(z,W(z,S(z,2n)))$ $\ \ \ by\ identity1$
$=W(z,W(z,pr_1+1))$
$=W(z,pr_2)$
$=pr+1=(z-1)r+1$
For some $r,r_1,r_2\in\mathbb{Z}$
I believe $r$ is always $1$ for all $z>2n+2$, that's my question.
Related questions on MSE
To count such $p$ which $p\nmid S(p,2n)$
Special observation on prime number and π(n)
Same question
 A: Define $X_a$ be the set as, $\{2,3,...,a-1,a\}$
let $D(b,m)$ be the sum of the base-$b$ digits of $m$.
Define $f(a,k)=\frac{D(a,a^{k+1}-S(a,k))}{a-1}$
Theorem:
Given $a\in \mathbb{Z}_{\ge 4}$ and $m\in \mathbb{Z}_{\ge 1}$,  If $a-1\mid S(a-1,2m)$ and $a-1>2m+1$ then $(f(a,2m))_a\in X_a$
incomplete Proof: This proof is incomplete to show $1\notin f(a,2m)$ but I assumed here it's true.
Clearly, we have $(a-1)|S(a-1,2m)$ iff $(a-1)|D(a,S(a-1,2m))$.
Let $q:=\frac{D(a,S(a-1,2m))}{a-1}$. Since for $a\geq 4$ and $m\geq 1$,
$S(a-1,2m) < (a-1)a^{2m}$ and $S(a,2m) = S(a-1,2m) + a^{2m}$, we have $D(a,S(a,2m)) = 1+q(a-1)$. Then
$$f(a,2m) = \frac{D(a,a^{2m+1} - a^{2m} - S(a-1,2m))}{a-1} \le 2m+1-q.$$
Since $2m+1-q<2m+1<a-1$, we conclude that $(f(a,2m))_a$ forms a single digit $2m+1-q\geq 2$, and thus $(f(a,2m))_a\in X_a$.
Corollary 1: $W(a+1,ax+1)=a$ if $x\in X_{a+1}$
Corollary 2: If $ p|S(p,2n)$ and $p\ge 2n+1$ Then $W(z,W(z,S(z,2n)))=z$
Proof
For $z>2n+2$
See $S(z,2n)=pr_1+1$ For some $r_1\in\mathbb{Z}$
$\implies W(z,W(z,S(z,2n)))$ $\ \ \ by\ identity1$
$=W(z,W(z,pr_1+1)) $
$=W(z,pr_2)$
Here by theorem 2 $\implies 2\le r_2 < p$
Hence $W(z,pr_2)=p+1=z$

Reference
Show that, If $a-1\mid S(a-1,2m)$ and $a-1>2m+1$ then $(f(a,2m))_a\in X_a$
