Problem related to inequality of sum of digits of power sum

Question

Let $D$ be the function define as $D(b,n)$ be the sum of the base-$b$ digits of $n$.

Example: $D(2,7)=3$ means $7=(111)_2\implies D(2,7)=1+1+1=3$

Define $S(a,m)=1^m+2^m+3^m+...+a^m$ where $a,m\in\mathbb{Z}_+$

Can it be shown that

For $a,m>1$,

$(a-1)m>D(a,S(a-1,m))$?

More on observation

$(1)\space (a-1)m=D(a,S(a-1,m)) \iff m=1,a\equiv0\pmod2$

$(2)$ if $m\ge 5,a\ge3$ then $D(a,S(a-1,m))>a-1$

Can someone please give me reference to understand this pattern, thanks.

Source code Pari/GP

for(m=1,50,for(a=2,100,if(sumdigits(sum(i=1,a-1,i^m),a)>=(a-1)*m,print([m,a,sumdigits(sum(i=1,a-1,i^m),a)]))))

Note: For $a,m>1$

● $a^m<S(a,m)<a^{m+1}$

● $1\le D(a,S(a,m))\le(a-1)(m+1)$

● $D(a,S(a,m))=1+D(a,S(a-1,m))$proof

●maybe this post helpful https://math.stackexchange.com/q/3595166/647719

The question was posted in MSE(10/1/20) but no answer hence posting in MO check here

Answer 1

This is surely the case if $m\geq a-1$, when we have $$S(a-1,m)<\int_1^a x^m\ dx<\frac{a^{m+1}}{m+1}\leq a^m.$$