Proving that the limit of $\root n \of {{a_1}^n + {a_2}^n + ... + {a_k}^n}$ is $\max(a_1,...a_k)$

$$\sqrt[n]{a_{\max}^n} \leq \sqrt[n]{a^n_1 + \dots + a_k^n} \leq \sqrt[n]{k \cdot a_{\max}^n}$$

Let's suppose, without any loss of generality, that $a_1 = \max \left\{ {{a_1},{a_2}...{a_k}} \right\}$. Then:

$$ \root n \of {{a_1}^n + {a_2}^n + ... + {a_k}^n} = \root n \of {a_1^n \left (1+\left (\dfrac{a_2}{a_1} \right )^n + ... + \left (\dfrac{a_k}{a_1} \right )^n \right )} \leq \root n \of {{a_1}^n \cdot k} = a_1 \root n \of {k}$$

Since $\dfrac{a_i}{a_1} \leq 1$, $i=1, ..., k$.

Therefore, $\lim_{n \to \infty} \root n \of {{a_1}^n + {a_2}^n + ... + {a_k}^n} = a_1$