$$\begin{aligned} a_1^k + \frac{1}{a_1^k} &= a_1^k + \frac{1}{n^{2k} a_1^k} + \cdots + \frac{1}{n^{2k} a_n^k} \\ &\ge (n^{2k} + 1) \left( \frac{a_1^k}{(n^{2k} a_1^{2k})^{n^{2k}}} \right)^{\frac{1}{n^{2k}+1}} \\ &= (n^k + \frac{1}{n^k}) \left( \frac{1}{n a_1} \right)^{\frac{k(n^{2k}-1)}{n^{2k}+1}}. \end{aligned}$$

$$\begin{aligned} a_2^k + \frac{1}{a_2^k} &\ge (n^k + \frac{1}{n^k}) \left(\frac{1}{n a_2}\right)^{\frac{k(n^{2k}-1)}{n^{2k}+1}}, \\ \vdots & \\ a_n^k + \frac{1}{a_n^k} &\ge (n^k + \frac{1}{n^k}) \left(\frac{1}{n a_n}\right)^{\frac{k(n^{2k}-1)}{n^{2k}+1}} \end{aligned}$$

Similarly obtained

Multiplying the $n$ inequalities together, we get
$$\begin{aligned} & (a_1^k + \frac{1}{a_1^k})(a_2^k + \frac{1}{a_2^k})\cdots(a_n^k + \frac{1}{a_n^k}) \\ & \ge (n^k + \frac{1}{n^k})^n \left(\frac{1}{n^n a_1 \cdots a_n}\right)^{\frac{k(n^{2k}-1)}{n^{2k}}} \end{aligned}$$

Since $a_i > 0, i = 1, \dots, n, \sum_{i=1}^n a_i = 1$, then
$$a_1 \cdots a_n \le \left( \frac{1}{n} \sum_{i=1}^n a_i \right)^n = \frac{1}{n^n},$$
then
$$\frac{1}{n^n a_1 \cdots a_n} \ge 1.$$
Also, since
$$\frac{k(n^{2k} - 1)}{n^{2k} + 1} > 0, \text{ therefore, }$$
$$\left(\frac{1}{n^n a_1 \cdots a_n}\right)^{\frac{k(n^{2k})}{n^{2k}+1}} \ge 1$$
Hence $(a_1^k + \frac{1}{a_1^k})(a_2^k + \frac{1}{a_2^k}) \cdots (a_n^k + \frac{1}{a_n^k}) \ge (n^k + \frac{1}{n^k})^n$