You can use the intuitive rules you know about derivatives of sums, but if you want a proof that is very explicit:
$$
G(x) = \bigg( \sum_{k=-\infty}^{\infty} g_k(x) \bigg)^n = \sum_{k_1,\dots,k_n| k_i \in \mathbb{Z}} \prod_{i = 1}^n g_{k_i}(x).
$$
The last summation is over $n$-tuples $(k_1, \dots, k_n)$ of integers. (Keep in mind that, if $n = 2$, then $(1,1)$, for example, would appear twice.) Let's assume that $G, g$ are such that the 1st and 2nd derivatives exist. Then
$$
G'(x) = \sum_{k_1,\dots,k_n| k_i \in \mathbb{Z}} \sum_{i=1}^n g'_{k_i}(x) \prod_{j \neq i}^n g_{k_j}(x),
$$
and
$$
G''(x) = \sum_{k_1,\dots,k_n| k_i \in \mathbb{Z}} \left( \sum_{i=1}^n g''_{k_i}(x) \prod_{j \neq i}^n g_{k_j}(x) + \sum_{i=1}^n \sum_{j\neq i}^n g'_{k_i}(x) g'_{k_j}(x) \prod_{l \neq i, j}^n g_{k_l}(x)\right).
$$
This does not look very helpful, but we can rewrite the expressions a bit:
$$
G'(x) = \sum_{i=1}^n \sum_{k_1,\dots,k_n| k_i \in \mathbb{Z}} g'_{k_i}(x) \prod_{j \neq i}^n g_{k_j}(x) = n \sum_{k_1 \in \mathbb{Z}} \sum_{k_2,\dots,k_n| k_i \in \mathbb{Z}} g'_{k_1}(x) \prod_{j = 2}^n g_{k_j}(x).
$$
This second equality comes from symmetry. The terms of the outside sum have the same value. It doesn't matter which of the $k_i$ is picked as $k_1$ for the derivative. Doing two more steps,
$$
G'(x) = n \sum_{k_1 \in \mathbb{Z}} g'_{k_1}(x) \left( \sum_{k_2,\dots,k_n| k_i \in \mathbb{Z}}\prod_{j = 2}^n g_{k_j}(x)\right) = n \sum_{k_1 \in \mathbb{Z}} g'_{k_1}(x) \left( \sum_{k = -\infty}^{\infty}g_k(x)\right)^{n-1}
$$
$$ = n \left( \sum_{i = -\infty}^{\infty} g'_i(x) \right) \left( \sum_{k = -\infty}^{\infty}g_k(x)\right)^{n-1}.
$$
Similarly, we can rewrite for $G''(x)$. The first part is processed similarly to the above. The other requires a bit more care.
$$
G''(x) = \sum_{k_1,\dots,k_n| k_i \in \mathbb{Z}} \sum_{i=1}^n g''_{k_i}(x) \prod_{j \neq i}^n g_{k_j}(x) + \sum_{k_1,\dots,k_n| k_i \in \mathbb{Z}}\sum_{i=1}^n \sum_{j\neq i}^n g'_{k_i}(x) g'_{k_j}(x) \prod_{l \neq i, j}^n g_{k_l}(x)
$$
$$
= n \left( \sum_{i = -\infty}^{\infty} g''_i(x) \right) \left( \sum_{k = -\infty}^{\infty}g_k(x)\right)^{n-1} + \sum_{i=1}^n \sum_{j\neq i}^n \sum_{k_1,\dots,k_n| k_i \in \mathbb{Z}} g'_{k_i}(x) g'_{k_j}(x) \prod_{l \neq i, j}^n g_{k_l}(x)
$$
$$
= n \left( \sum_{i = -\infty}^{\infty} g''_i(x) \right) \left( \sum_{k = -\infty}^{\infty}g_k(x)\right)^{n-1} +
n(n-1) \sum_{k_1,k_2 \in \mathbb{Z}} \sum_{k_3,\dots,k_n| k_i \in \mathbb{Z}} g'_{k_1}(x) g'_{k_2}(x) \prod_{l = 3}^n g_{k_l}(x).
$$
This last equality comes again from symmetry. The terms corresponding to the choice $i, j$ have the same value. There are $n(n-1)$ to choose the ordered pair $i, j$, with $j \neq i$ from $\{1, \dots, n\}$. Continuing,
$$
G''(x) = n \left( \sum_{i = -\infty}^{\infty} g''_i(x) \right) \left( \sum_{k = -\infty}^{\infty}g_k(x)\right)^{n-1} +
n(n-1)\left( \sum_{k_1,k_2 \in \mathbb{Z}} g'_{k_1}(x) g'_{k_2}(x) \right) \left( \sum_{k_3,\dots,k_n| k_i \in \mathbb{Z}} \prod_{l = 3}^n g_{k_l}(x) \right)
$$
$$
= n \left( \sum_{i = -\infty}^{\infty} g''_i(x) \right) \left( \sum_{k = -\infty}^{\infty}g_k(x)\right)^{n-1} +
n(n-1)\left( \sum_{i,j = -\infty}^{\infty} g'_{i}(x) g'_{j}(x) \right) \left( \sum_{k = -\infty}^{\infty}g_k(x)\right)^{n-2}
$$
$$
= n \left( \sum_{i = -\infty}^{\infty} g''_i(x) \right) \left( \sum_{k = -\infty}^{\infty}g_k(x)\right)^{n-1} +
n(n-1)\left( \sum_{i = -\infty}^{\infty} g'_{i}(x) \right) \left( \sum_{j = -\infty}^{\infty} g'_{j}(x) \right) \left( \sum_{k = -\infty}^{\infty}g_k(x)\right)^{n-2}
$$
$$
= n \left( \sum_{i = -\infty}^{\infty} g''_i(x) \right) \left( \sum_{k = -\infty}^{\infty}g_k(x)\right)^{n-1} +
\frac{n(n-1)}{2}\left( \sum_{i < j| i, j \in \mathbb{Z}} g'_{i}(x) g'_{j}(x) \right) \left( \sum_{k = -\infty}^{\infty}g_k(x)\right)^{n-2}.
$$
There are multiple ways of representing the last result, but they follow the chain rule intuition.
