<s>[INST] Proving $r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1} < a^k$ by INDUCTION. Let $a$ be a natural number $>1$. For all integers $r_0, r_1, \dots, r_{n-1}$ with $0\leq r_{j} < a$, then
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{n-1}a^{n-1} < a^n.
\end{eqnarray}

Consider the base case for $n_0=2$. Thus $r_0+r_1a<a^2$, which is not so obviously true, but given that $0\leq r_{j} < a$ and the conditions on both $a$ and the $r_j$'s, then the maximum value that $r_0+r_1a$ can attain is when $r_0=r_1=(a-1)$. Thus, we have
\begin{eqnarray}
r_0+r_1a & = & (a-1)+(a-1)a\\
& = & (a-1)(a+1)\\
& = & a^2-1,
\end{eqnarray}
which clearly suffices to show that the inequality $r_0+r_1a<a^2$ holds, namely $a^2-1<a^2$. Now, let $k$ be some number $>2$. By assumption
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1} < a^k,
\end{eqnarray}
with the same conditions as are listed in the theorem above. However, we must show that
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1}+r_ka^k < a^{k+1}.
\end{eqnarray}
To continue, let us add $r_ka^k$ to both sides of the inequality we made by assumption:
\begin{eqnarray}
r_0+r_1a+r_2a^2+\cdots+r_{k-1}a^{k-1}+r_ka^k & < & a^k+r_ka^k\\
& = & a^k(r_k+1)\cdots \text{Where to now? :-(}
\end{eqnarray}
 [/INST] Let $R$ be $\max(r_i)\implies R\le a-1$ as $r_i$s are integers  (As Aryabhata commented)
So, $$r_0+r_1a+r_2a^2+\cdots+r_{n-1}a^{n-1}$$
$$\le R\cdot(1+a+a^2+\cdots+a^{n-1})$$
$$\le(a-1)\cdot\left(\frac{a^n-1}{a-1}\right)=a^n-1\text{ as }a\ne1 \text{ as }a>1$$
$$<a^n$$
</s>