Prove for every $i∈\Bbb Z∪{0}$ that $N(b^i)Let $D$ be a Euclidean domain with euclidean valuation $N$ . Suppose $b∈D$ is neither zero nor a unit. Prove for every $i∈\Bbb Z∪{0}$ that $N(b^i)<N(b^{i+1})$ .
Euclidean valuation is a function:
$N:D→\Bbb N$ s.t
a) $∀α∈D$ $β∈D\setminus\{0\}$ $\exists q,r \in D$ s.t $a=bq+r$ where either $r=0$ or $N(r)<N(b)$
b) $N(αβ)\geq N(α)$ ; $∀α,β∈D\setminus\{0\}$.
Basically, I am struggling to prove that $N(b)>1$ if b is neither a unit nor zero.
 A: We prove the following lemma:

Lemma 1: Let $D$ be an Euclidean domain and assume that $b\in D$ is not a unit and not zero. Then $N(b)<N(ab)$ for all $a\in D\backslash\lbrace 0\rbrace$.

Proof: Assume by way of contradiction that $N(ab)=N(b)$ (we know that $N(ab)\geq N(b)$ by property b) mentioned in your post). We can find $q,r\in D$ such that
$$a=(ab)q+r\hspace{5mm}\text{with}\hspace{5mm}r=0\hspace{2.5mm}\text{or}\hspace{2.5mm}N(r)<N(q),$$
because $ab\neq 0$. Now, note that we can write
$$0=a(1-bq)+(-r),$$
so either $r=0$ or $N(-r)<N(a)$ by the property a). Note that $N(-r)=N(r)$. Indeed, we have
$$N(r)\leq N(-1\cdot r)\leq N(-1\cdot -1\cdot r)=N(r).$$
Thus either $r=0$ or $N(r)<N(a)$. Since $b$ is not a unit we have that $1-bq\neq 0$, whence $a(1-bq)\neq 0$ because $D$ is an integral domain and $a\in D\backslash\lbrace 0\rbrace$. We now see that $r\neq 0$. Indeed, by the equation above we have
$$0=a(1-bq)+(-r)\ \Rightarrow \ r=a(1-bq)\neq 0,$$
so we conclude that
$$N(a(1-bq))=N(r)<N(a).$$
By property b) we have
$$N(a)\leq N(a(1-bq))$$
because $1-bq\neq 0$, but this is a contradiction and we conclude that $N(ab)\neq N(b)$, i.e. $N(b)<N(ab)$.
$$\tag*{$\blacksquare$}$$
Now, prove that $\beta$ is a unit in $D$ if and only if $N(\beta)=N(1)$ and apply the lemma to obtain your desired result.
