real number and decimal expansions
$$ \newcommand{\bb}[1]{\left( #1 \right)} \newcommand{\f}[1]{\left\lfloor #1 \right\rfloor} $$ Write $x_0 := N$. Note that your expression becomes the following: $$ x_i = \f{10^i\bb{x - \sum_{j=0}^{i-1}\frac{x_j}{10^j}}} $$
For (1), we can make use of the following lemma:
Lemma: For any $k \in \mathbb{N}$, we have: $$ \sum_{i=0}^k \frac{x_i}{10^i} = \frac{\f{10^{k}x}}{10^{k}} $$
Proof. We prove by induction. The case is clear for $k = 0$, as by definition $x_0 = \f{x}$. Now suppose $\sum_{i=0}^k \frac{x_i}{10^i} = \frac{\f{10^kx}}{10^k}$. Then: \begin{align*} \sum_{i=0}^{k+1} \frac{x_i}{10^i} &= \frac{\f{10^kx}}{10^k} + \frac{x_{k+1}}{10^{k+1}} \\ &= \frac{\f{10^kx}}{10^k} + \frac{1}{10^{k+1}}\f{10^{k+1}\bb{x - \sum_{j=0}^{k}\frac{x_j}{10^j}}} \\ &= \frac{\f{10^kx}}{10^k} + \frac{1}{10^{k+1}}\f{10^{k+1}x - 10^{k+1}\frac{\f{10^kx}}{10^k}} \\ &= \frac{\f{10^kx}}{10^k} + \frac{1}{10^{k+1}}\f{10^{k+1}x - \underbrace{10\f{10^kx}}_\text{integer}} \\ &= \frac{\f{10^kx}}{10^k} + \frac{1}{10^{k+1}}\bb{\f{10^{k+1}x} - 10\f{10^kx}} \\ &= \frac{\f{10^kx}}{10^k} + \frac{\f{10^{k+1}x}}{10^{k+1}} - \frac{\f{10^kx}}{10^k} \\ &= \frac{\f{10^{k+1}x}}{10^{k+1}} \end{align*} Now, it's simple to prove that $0 \leq x_i \leq 9$. We observe that: \begin{align*} x_i = \f{10^ix - 10^i\frac{\f{10^{i-1}x}}{10^{i-1}}} = \f{10^ix - 10\f{10^{i-1}x}} = \f{10\bb{10^{i-1}x - \f{10^{i-1}x}}} \end{align*} We know that for any integer $n$, $0 \leq n - \f{n} < 1$. Thus: \begin{align*} 0 \leq 10^{i-1}x - \f{10^{i-1}x} < 1 &\implies 0 \leq 10\bb{10^{i-1}x - \f{10^{i-1}x}} < 10 \\ &\implies 0 \leq \f{10\bb{10^{i-1}x - \f{10^{i-1}x}}} \leq 9 \end{align*} So $0 \leq x_i \leq 9$.
For (2), we shall show that there is no $M \in \mathbb{Z}^+$ such that for $i > M$, $x_i = 9$. Suppose such an $M \geq 1$ exists, and suppose $x_{M} = n$. We observe that for $M' > M$: \begin{align*} 10^{M}\bb{x - \sum_{j=0}^{M-1} \frac{x_j}{10^j}} - (n + 1) &= 10^{M}\bb{x - \sum_{j=0}^{M-1} \frac{x_j}{10^j}} - 1 - n \\ &\geq^* 10^{M}\bb{\sum_{j=0}^{M'}\frac{x_j}{10^j} - \sum_{j=0}^{M-1} \frac{x_j}{10^j}} - 1 - n \\ &= 10^M\sum_{j=M}^{M'} \frac{x_j}{10^j} - 1 - n\\ &= 10^M\sum_{j=M+1}^{M'} \frac{x_j}{10^j} - 1 \\ &= 10^M\sum_{j=M+1}^{M'} \frac{9}{10^j} - 1 \\ &= 10^M\frac{\frac{9}{10^{M+1}}\bb{1 - \frac{1}{10^{M' - M}}}}{1 - \frac{1}{10}} - 1\\ &= - \frac{1}{10^{M' - M}} \end{align*} We can let $M' \to +\infty$, and we have that $10^{M}\bb{x - \sum_{j=1}^{M-1} \frac{x_j}{10^j}} - (n + 1) \geq 0$. Thus: $$ \f{10^{M}\bb{x - \sum_{j=0}^{M-1} \frac{x_j}{10^j}} - (n + 1)} \geq 0 \implies x_M \geq n + 1 $$ which contradicts that $x_M = n$. Note that the starred inequality can be easily proven as follows: $$ x - \sum_{i=0}^{M'} \frac{x_i}{10^i} = x - \frac{\f{10^{M'}x}}{10^{M'}} \geq x - \frac{10^{M'}x}{10^{M'}} = 0 $$