Prove that there is only one unique base b representation of any natural number.

Let $N$ be a natural number. You want to write it as $$ N = a_{0} + a_{1} b + a_{2} b^{2} + \dots + a_{k} b^{k} $$ for a suitable $k$, and $0 \le a_{i} < b$.

Now rewrite the above as $$ \begin{cases} N = a_{0} + q b,\\ 0 \le a_{0} < b \end{cases}$$ to see that $a_{0}$ is uniquely determined as the remainder of the division of $N$ by $b$.

Now consider $$ q = a_{1} + a_{2} b + \dots + a_{k} b^{k-1} $$ and repeat, that is, use induction.

Below we show that uniqueness of radix rep is a special case of the Rational Root Test.

If $\,g(x) = \sum g_i x^i$ is a polynomial with integer coefficients $\,g_i\,$ such that $\,0\le g_i < b\,$ and $\,g(b) = n\,$ then we call $\,(g,b)\,$ a radix $\,b\,$ representation of $\,n.\,$ It is unique: $ $ if $\,n\,$ has another rep $\,(h,b),\,$ with $\,g(x) \ne h(x),\,$ then $\,f(x)= g(x)-h(x)\ne 0\,$ has root $\,b\,$ but all coefficients $\,\color{#c00}{|f_i| < b},\,$ contra the below slight generalization of: $ $ integer roots of integer polynomials $\,f(x)\in\Bbb Z[x]\,$ divide its constant term $\,f(0)\,$ [an obvious special case of the Rational Root Test].

Theorem $\ $ If $\,f(x) = x^k(\color{#0a0}{f_0}\!+f_1 x +\cdots + f_n x^n)=x^k\bar f(x)\,$ is a polynomial with integer coefficients $\,f_i\,$ and with $\,\color{#0a0}{f_0\ne 0}\,$ then an integer root $\,b\ne 0\,$ satisfies $\,b\mid f_0,\,$ so $\,\color{#c00}{|b| \le |f_0|}$

Proof $\ \ 0 = f(b) = b^k \bar f(b)\,\overset{\large b\,\ne\, 0}\Rightarrow\, 0 = \bar f(b),\,$ so, subtracting $\,f_0$ from both sides yields $$-f_0 =\, b\,(f_1\!+f_2 b+\,\cdots+f_n b^{n-1})\, \Rightarrow\,b\mid f_0 \underset{\large \color{#0a0}{f_0\,\ne\, 0}}\Longrightarrow\, |b| \le |f_0|\qquad {\bf QED}\qquad\quad$$