Given an element $a\in\mathcal A$ which is self-adjoint, it is easy to prove that $\|a^2\|=\|a\|^2$, from $\|a^2\|=\|aa^*\|=\|a\|^2$. This can be generalised to say that $\|a^{2^k}\|=\|a\|^{2^k}$.
What about the other powers? Can you have for example $\|a^3\|\neq \|a\|^3$ for a self-adjoint $a$? If so, what is an example of an operator that does this?
This doesn't seem to be possible in a finite-dimensional Hilbert space, as if $A$ is a self-adjoint operator then $A^n=\sum_k \lambda_k^n u_k u_k^*$ with $\lambda_k\in\mathbb R$ its eigenvalues, and thus $$\|A^n\|=\max\{\lambda_k^n\}=(\max\lambda_k)^n=\|A\|^n.$$
If $a$ is self-adjoint, we have $||a||=r(a)$, where $r(a)$ denotes the spctral radius of $a$. This follows from $\|a^{2^k}\|=\|a\|^{2^k}$ for all $k$.
On the other hand, $r(a) = \inf_{n}||a^n||^{1/n}.$
Hence $||a|| \le ||a^n||^{1/n}$ for all $n$. Therefore $||a||^n \le ||a^n||$ for all $n$. Since the norm is multiplictive, we have $||a^n|| \le ||a||^n$.
Conclusion: $||a^n||=||a||^n$ for all $n$.