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STATEMENT: Let $A$ be a Banach algebra, then for every $x\in A$ we have $$\lim_{n\rightarrow\infty}\|x^n\|^{1/n}=r(x)$$

Proof: We know that $r(x)\leq \lim \inf_n\|x^n\|^{1/n}$, so it suffices to prove that $$\limsup_{n\rightarrow \infty}\|x^n\|^{1/n}\leq r(x)$$ We need only consider the case $x\neq 0$. To prove our formula choose $\lambda \in \mathbb{C}$ satisfying $|\lambda|<1/r(x)$. We claim that the sequence $\left\{(\lambda x)^n:n-1,2,...\right\}$ is bounded.

Indeed, by the Banach-Steinhaus theorem it suffices to show thatfor every bounded linear function$\rho$ on $A$ we have the following
$$|\rho(x^n)\lambda^n|=|\rho((x\lambda)^n)|\leq M_p<\infty$$

Question: I don't see why Banach-Steinhaus theorem would be sufficient to show that the sequence is bounded. This proof is from A Short Course in Spectral Theory by William Arveson.

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We can apply the Banach-Steinhaus theorem (also known as the Uniform boundedness principle) to your situation as follows.

Let $X$ be a Banach space. Then we have the evaluation map from $X$ to its bidual $X''$ $$ \varphi:X\rightarrow X'', x \rightarrow \varphi(x) $$ with $$ (\varphi(x))(f) = f(x)\qquad for\ x\in X\ and\ f\in X'. $$ It is "well known" that $\varphi$ is a linear isometry (see for example here).

Now, in your situation, we are given the sequence $((\lambda x)^n)$ in $X.$ Applying $\varphi,$ we get a sequence $(\varphi((\lambda x)^n))$ in $X''.$ Suppose now we are given that $$ \forall \rho \in X'\exists M_\rho < \infty: |\rho((\lambda x)^n)| = |(\varphi((\lambda x)^n))(\rho)| \leq M_\rho, $$ i.e. the sequence $(\varphi((\lambda x)^n))$ of continuous linear operators from $X'$ to $\mathbb C$ is pointwise bounded. Then Banach-Steinhaus tells us that this sequence is norm-bounded, in symbols $$ \exists M < \infty \forall n\in \mathbb N: \|\varphi((\lambda x)^n)\|_{X''}= \|(\lambda x)^n)\|_X \leq M. $$ Here we have used that $\varphi$ is an isometry.

Thus we have shown that the original sequence $((\lambda x)^n)$ is bounded, as desired.

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