Consider that for a norm we have the submultiplicative property, then $\|a^n\|\leq \|a\|^n $.
If we have a value $b$ such that $b \leq \|a\|$, then how can it be that for any $\varepsilon>0$: $$\|a^n\| \leq (b+\varepsilon)^n$$ for a $n$ sufficiently big?
Isuppose that the reasoning for that is that $\|a^n\|-\|a\|^n$ is increasing with each passing $n$ (not actually sure how to prove that) and therefore for a big enough $n$ one can make the value $(b+\varepsilon)^n$ be bigger than $\|a^n\|$, even if it is smaller than $\|a\|^n$?
This question actually comes from a obscure passage in a proof that the $C^*$ condition $\|a^*a\| = \|a^*\|\|a\|$ is equivalent to the $B^*$ condition $\|a^*a\| = \|a\|^2$ for $C^*$-algebras, let's call the $C^*$-algebra in question $\mathscr{A}$.
Considering that $r(a) := \sup\limits_{\lambda \in \sigma(a)} |\lambda|\,\,\,$ (what I had previously called $b\,$)$\,\,\,$ is the spectral radius which has the property that $r(a) \leq \|a\|,\,\forall a \in \mathscr{A}$, then the passage says that:
$``$For every $\varepsilon >0$ there is a integer $n$ such that: $$\|(x^*y)^{2^n}\| \leq (r(x^*y)^2 +\varepsilon)^{2^{n-1}}\,\,\,\text{and}\,\,\,\,\,\, \|(y^*x)^{2^n}\| \leq (r(y^*x)^2 +\varepsilon)^{2^{n-1}} \,\,\,\,\,"$$
This looked false to me the first time I looked at this since the exponentiation by $2^n$ is a monotonously increasing function and so I could not see how $(r(x^*y)^2 +\varepsilon)$ could gain from $\|x^*y\|$ with a value $r(x^*y)$ that is smaller than $\|x^*y\|$ and a arbitrary value (that could keep being smaller when summed) $\varepsilon>0$, but the submultiplicative property of the norm seemsto factor in strongly in this assertion (No other comment is made in the refernce about this inequality beyound what is in the quotation marks). I have tried to generalise this property in the question since I don't think this inequality is necessarily valid only for the specific form presented between the quotation marks.
The claim is true when $r(a) \le b$, which is satisfied in your example. Indeed, by Gelfand's formula we know that $$r(a) = \lim_{n\to\infty} \|a^n\|^{\frac1n}.$$ Therefore, for any $\varepsilon > 0$ there exists $n_\varepsilon \in \Bbb{N}$ such that for all $n \ge n_\varepsilon$ we have $$\|a^n\|^{\frac1n} \le r(a) + \varepsilon \le b + \varepsilon \implies \|a^n\| \le (b+\varepsilon)^n.$$