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I just want to give some other proof to your second question. Let $SS(m^n)=\lbrace S(m^n) ; n\geq1\rbrace$. We show that this set is unbounded. First a Lemma:

$Lemma:$ Let $a$ be any arbitrary integer unless your trivial cases. For any arbitrary integer $b$, there is an integer $t$ such that $a^t$ begins with $b$.

Now by contrary, suppose $Max$ $SS(m^n)=M$, and just look to this sequence of integer:

$\overline {M1}, \overline {M11}, \ldots$

this sequence is infinite and by previous lemma, for each of them you must have a suitable $t_i$, $i$ is the position of each number in the above sequence, that $m^{t_i}$ begins with $i$-th term in the sequences. Obviously, each of them has the digit sum greater than $M$ and it is a contradiction. This completes the proof.

I just want to give some other proof to your second question. Let $SS(m^n)=\lbrace S(m^n) ; n\geq1\rbrace$. We show that this set is unbounded. First a Lemma:

$Lemma:$ Let $a$ be any arbitrary integer unless your trivial cases. For any arbitrary integer $b$, there is an integer $t$ such that $a^t$ begins with $b$.

Now by contrary, suppose $Max$ $SS(m^{n_1})=M$ for some $n_1\in N$, and just look to this sequence of integer:

$\overline {m^{n_1}1}, \overline {m^{n_1}11}, \overline {m^{n_1}111}, \ldots$

this sequence is infinite and by previous lemma, for each of them you must have a suitable $t_i$, $i$ is the position of each number in the above sequence, that $m^{t_i}$ begins with $i$-th term in the sequences. Obviously, each of them has the digit sum greater than $M$ and it is a contradiction. This completes the proof.

I just want to give some other proof to your second question. Let $SS(m^n)=\lbrace m^n ; n\geq1\rbrace$. We show that this set is unbounded. First a Lemma:

$Lemma:$ Let $a$ be any arbitrary integer unless your trivial cases. For any arbitrary integer $b$, there is an integer $t$ such that $a^t$ begins with $b$.

Now suppose $Max$ $SS(m^n)=M$, and just look to this sequence of integer:

$\overline {M1}, \overline {M11}, \ldots$

this sequence is infinite and by previous lemma, for each of them you must have a suitable $t_i$, $i$ is the position of each number in the above sequence, that $m^{t_i}$ begins with $i$-th term in the sequences. Obviously, each of them has the digit sum greater than $M$ and it is a contradiction. This completes the proof.

I just want to give some other proof to your second question. Let $SS(m^n)=\lbrace S(m^n) ; n\geq1\rbrace$. We show that this set is unbounded. First a Lemma:

$Lemma:$ Let $a$ be any arbitrary integer unless your trivial cases. For any arbitrary integer $b$, there is an integer $t$ such that $a^t$ begins with $b$.

Now by contrary, suppose $Max$ $SS(m^n)=M$, and just look to this sequence of integer:

$\overline {M1}, \overline {M11}, \ldots$

this sequence is infinite and by previous lemma, for each of them you must have a suitable $t_i$, $i$ is the position of each number in the above sequence, that $m^{t_i}$ begins with $i$-th term in the sequences. Obviously, each of them has the digit sum greater than $M$ and it is a contradiction. This completes the proof.

I just want to give some other proof to your second question. Let $SS(m^n)=\lbrace m^n ; n\geq1\rbrace$. We show that this set is unbounded. First a Lemma:

$Lemma:$ Let $a$ be any arbitrary integer unless your trivial cases. For any arbitrary integer $b$, there is an integer $t$ such that $a^t$ begins with $b$.

Now suppose $Max$ $SS(m^n)=M$, and just look to this sequence of integer:

$\overline {M1}, \overline {M11}, \ldots$

this sequence is infinite and by previous lemma, for each of them you must have a suitable $t_i$, $i$ is the position of each number in the above sequence, that $m^n$ begins with $i$-th term in the sequences. Obviously, each of them has the digit sum greater than $M$ and it is a contradiction. This completes the proof.

I just want to give some other proof to your second question. Let $SS(m^n)=\lbrace m^n ; n\geq1\rbrace$. We show that this set is unbounded. First a Lemma:

$Lemma:$ Let $a$ be any arbitrary integer unless your trivial cases. For any arbitrary integer $b$, there is an integer $t$ such that $a^t$ begins with $b$.

Now suppose $Max$ $SS(m^n)=M$, and just look to this sequence of integer:

$\overline {M1}, \overline {M11}, \ldots$

this sequence is infinite and by previous lemma, for each of them you must have a suitable $t_i$, $i$ is the position of each number in the above sequence, that $m^{t_i}$ begins with $i$-th term in the sequences. Obviously, each of them has the digit sum greater than $M$ and it is a contradiction. This completes the proof.