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Proving with exponents
Proving identities of any sort has never been my strongpoint. No, this is not trig identities. I have attempted to figure out this problem multiple times with rearranging and such and I have yet to be able to prove it. The answers do not show any steps or anything, it just says: True. The question looks like:
[3^(2n) 27^(5n+1) 9^(6n)] / 81^(4n7) = [ 9^(7n) 3^(8n+28)] / 27^(3n1)
If anyone could explain to me their process of solving this I would be grateful.

I get
$\displaystyle \displaystyle \frac{3^{2n} 27^{5n+1} 9^{6n}}{ 81^{4n7} }$
$\displaystyle \displaystyle \frac{3^{2n} (3^3)^{5n+1} (3^2)^{6n}}{ (3^4)^{4n7} }$
$\displaystyle \displaystyle \frac{3^{2n} 3^{3(5n+1)} 3^{2\times 6n}}{ 3^{4(4n7)} }$
$\displaystyle \displaystyle \frac{3^{2n} 3^{15n+3} 3^{12n}}{ 3^{16n28} }$
$\displaystyle \displaystyle \frac{3^{2n+(15n+3)+(12n)}}{ 3^{16n28} }$
$\displaystyle \displaystyle \frac{3^{29n+3}}{ 3^{16n28)} }$
$\displaystyle \displaystyle 3^{(29n+3)(16n28)}$
Finish it.