Let $k\in\mathbb{N}$ and $(G_i)_{1\leq i\leq k}$ a finite sequence of groups. I want to prove $$\prod_{i=1}^k[G_i,G_i]=\left[\prod_{i=1}^k G_i,\prod_{i=1}^kG_i\right].$$ For $k=2$, I am able to derive the equality $$\prod_{i=1}^2[G_i,G_i]=\left[\prod_{i=1}^2 G_i,\prod_{i=1}^2G_i\right].$$ However, assuming the equality for arbitrary $k\ge 2$, I am only able to obtain the isomorphism $$\prod_{i=1}^{k+1}[G_i,G_i]\cong\left[\prod_{i=1}^{k+1} G_i,\prod_{i=1}^{k+1}G_i\right].$$
Let's assume the relation for some $k\ge 2$. Let $I=\big\{(1,[1,k]),(2,\{k+1\})\big\}$. Then $$\prod_{i=1}^{k+1}[G_i,G_i]\cong\prod_{j=1}^{2}\prod_{i\in I_j}[G_i,G_i].$$ Note that $\prod_{i\in I_j}[G_i,G_i]=\left[\prod_{i\in I_j}G_i,\prod_{i\in I_j}G_i\right]$. Hence
$$\prod_{j=1}^{2}\prod_{i\in I_j}[G_i,G_i]=\prod_{j=1}^{2}\left[\prod_{i\in I_j}G_i,\prod_{i\in I_j}G_i\right]=\left[\prod_{j=1}^{2}\prod_{i\in I_j}G_i,\prod_{j=1}^{2}\prod_{i\in I_j}G_i\right]\cong\left[\prod_{i=1}^{k+1}G_i,\prod_{i=1}^{k+1}G_i\right].$$
Therefore, I only possess a mere isomorphism instead of required equality. Am I doing something wrong? Is this one of those situation where we simply treat the isomorphism as an equality?
The subgroup $[\prod G_i,\prod G_i]$ is generated by elements of the form $[(g_i),(h_i)]$.
Since $$\begin{align*} [(g_i),(h_i)] &= (g_i)^{-1}(h_i)^{-1}(g_i)(h_i)\\ &= (g_i^{-1})(h_i^{-1})(g_i)(h_i)\\ &= (g_i^{-1}h_i^{-1}g_ih_i)\\ &= ([g_i,h_i])\\ &\in \prod([G_i,G_i]), \end{align*}$$ we see that $[\prod G_i,\prod G_i]$ is a subgroup of $\prod[G_i,G_i]$.
Conversely, $\prod[G_i,G_i]$ is generated by elements of the form $$(e,e,\ldots,e,[g_j,h_j],e,\ldots,e)$$ which are trivial in all except perhaps for the $j$th coordinate, where it is a simple commutator. Now let $\mathbf{g},\mathbf{h}\in\prod G_i$ be the elements that are trivial in all but the $j$th coordinate, $(\mathbf{g})_j = g_j$, and $(\mathbf{h})_j = h_j$. Then it is straightforward to verify that $$[\mathbf{g},\mathbf{h}] = (e,e,\ldots,e,[g_j,h_j],e,\ldots,e).$$ Thus, every generator of $\prod[G_i,G_i]$ lies in $[\prod G_i,\prod G_i]$, so $\prod[G_i,G_i]$ is a subgroup of $[\prod G_i,\prod G_i]$. This proves the desired equality.
We strongly use the fact that it is a finite product in the second part. For an arbitrary product, $[\prod G_i,\prod G_i]$ is contained in $\prod[G_i,G_i]$, but equality may fail to hold.
If you want to prove it by induction, once you prove it for $k=2$ as an equalty, we have that $\prod_{i=1}^{k+1}G_i = (\prod_{i=1}^k G_k)\times G_{k+1}$. So we have $$\begin{align*} \left[\prod_{i=1}^{k+1}G_i,\prod_{i=1}^{k+1}G_i\right] &= \left[ \prod_{i=1}^kG_i\times G_{k+1},\prod_{i=1}^kG_i \times G_{k+1}\right] \\ &= \left[ \prod_{i=1}^k G_i,\prod_{i=1}^k G_i\right]\times [G_{k+1},G_{k+1}]\\ &= \prod_{i=1}^k[G_i,G_i] \times [G_{k+1},G_{k+1}]\\ &= \prod_{i=1}^{k+1}[G_i,G_i], \end{align*}$$ giving the desired equality. The second equality is from the $k=2$ case, and the third from the induction hypothesis.
Note that although technically the sets $(A_1\times A_2)\times A_3$, $A_1\times(A_2\times A_3)$, and $\prod_{i=1}^3 A_i$ are different (the first consists of ordered pairs, first entry an ordered pair, second entry in $A_3$; the second consists of ordered pairs, first entry in $A_1$, second entry an ordered pair; and the third set a collection of functions $f\colon\{1,2,3\}\to A_1\cup A_2\cup A_3$ with $f(i)\in A_i$), by convention we consider all three “the same set”; in fact, there is a canonical isomorphism between any two since the three sets satisfy the relevant universal property, and so we really do consider them “the same”, as they are isomorphic up to unique isomorphism. This is a bit of an abuse of notation, but it is an innocuous one. Your “isomorphism” is actually a canonical one, and so you can safely “identify” the two and claim equality.