The proof of SVD solver formula, $C^T C = V \Sigma^T \Sigma V^T$

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This formula could be used to compute matrix multiplication transpose

$${\displaystyle (\mathbf {AB} )^{\mathsf {T}}=\mathbf {B} ^{\mathsf {T}}\mathbf {A} ^{\mathsf {T}}}$$

This formula is used to elaborate matrix multiplication associativity

$${\displaystyle (\mathbf {AB} )\mathbf {C} =\mathbf {A} (\mathbf {BC} )}$$

This formula is used to compute svd of a matrix.

$$C = U \Sigma V^T$$

this MIT course puts them together and gives (equation_1)

$$C^T C = V \Sigma^T \Sigma V^T \tag 1$$

what is the detailed proof for equation_1?

first, what is the detailed procedure of $C^TC = (U \Sigma V^T)^TC$

how can I transfer this to $U^TU$?

$$C^T = V \Sigma^T U^T$$

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Using your first two equations, you can get

$$(ABC)^T = ((AB)C)^T = C^T(AB)^T = C^T B^T A^T$$

and then

$$C^T = V \Sigma^T U^T$$

and then

$$C^T C = V \Sigma^T U^T C = V \Sigma^T U^T U \Sigma V^T$$

since

$$U^TU = I$$

eventually you get

$$C^T C = V \Sigma^T U^T U \Sigma V^T = V \Sigma^T \Sigma V^T$$