Explicitness in numeral system 
Prove that for every $a\in\mathbb{N}$ there is one and only one way to express it in the system with base $\mathbb{N}\ni s>1$.

Seems classical, but I don't have any specific argument.

Suppose that some positive integer $N$ has two representations base $s$. Then we can write $$N=\sum_{k=0}^ma_ks^k=\sum_{k=0}^nb_ks^k\;,\tag{1}$$ where $0\le a_1,\dots,a_m,b_1,\dots,b_n\le s-1$ are integers, and without loss of generality we may assume that $m\le n$.
Suppose first that $m<n$. Then
$$\begin{align*}
\sum_{k=0}^ma_ks^k&\le\sum_{k=0}^m(s-1)s^k\\
&=\sum_{k=0}^ms^{k+1}-\sum_{k=0}^ms^k\\
&=\sum_{k=1}^{m+1}s^k-\sum_{k=0}^ms^k\\
&=s^{m+1}-1\\
&<s^n\\
&\le\sum_{k=0}^nb_ks^k\;,
\end{align*}$$
contradiction the assumption $(1)$. Thus, we assume that $m=n$.
Now make the further assumption that $N$ is the smallest positive integer with two different base $s$ representations. Without loss of generality assume that $a_n\le b_n$, and let $M=N-a_ns^n$. Then 
$$M=\sum_{k=0}^{n-1}a_ks^k=\sum_{k=0}^{n-1}b_ks^k+(b_n-a_n)s^n\;.\tag{2}$$
We just proved that this is impossible if $b_n-a_n\ne 0$, so $(2)$ must simplify to
$$M=\sum_{k=0}^{n-1}a_ks^k=\sum_{k=0}^{n-1}b_ks^k\;.$$
But then $M$ is a number with two base $s$ representations that is smaller than $N$, contradicting our choice of $N$. It follows that no such $N$ exists and hence that every positive integer has a unique representation base $s$.
