Convert recursive into explicit form

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Recursively define $f_n \in C^\infty (\mathbb{R}^n,\mathbb{R}^n)$ through

$f_2(y):=(y^1\cos y^2, y^1 \sin y^2)$ for $y=(y^1,y^2)\in \mathbb{R}^2$

and

$f_{n+1}(y)=(f_n(y')\sin y^{n+1},y^1 \cos y^{n+1})$ for $y=(y',y^{n+1})\in\mathbb{R}^n\times \mathbb{R}$

This exercise is a precursor to another one about manifolds and is from a book.

My solution:

$\displaystyle f_n(y)=y^1\left ( \prod_{j=3}^n\sin y^j(\sin y^2,\cos y^2), \bigoplus_{p=3}^n\cos y^p\prod_{j=p+1}^n \sin y^j\right )$

Is this correct? Just out of interest, is there any systematic way to get an explicit form from a recursive one?

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To remove the question from the list of unanswered.

Provided the recurrence holds for each $n\ge 2$, your answer is correct up to a swap of the first and second coordinates, provided $\bigoplus_{j=p}^n x_j$ means $(x_j,\dots, x_p)$, skipping inner brackets (for instance, assuming that $(x_1, (x_2, x_3))$ means $(x_1.x_2,x_3)$, and $\prod_{j=n+1}^n\sin y^j=1$.

As far as I know even for numbers instead of functions there is no systematic way to get an explicit form from a recursive one. But some particular cases can be solved by special methods, see, for instance, Wikipedia.