Kunfuuuu Power Play

LHS = $\huge {2^{3^{ \left( 2^{3^{2}} \right)^{0^{2^{3}}}}}}= \huge 2^{3^{2^{9 \times 0}}} = 2^{3^{9^{0}}}= \boxed{2^{3}}$ .

RHS = $\huge 2^{3^{1^{2^{3^{2^{0^{2^{3}}}}}}}} = 2^{3^{1}} = \boxed{2^{3}}$ .

As LHS = RHS, the given equation $\huge \color{#20A900}{2^{3^{ \left( 2^{3^{2}} \right)^{0^{2^{3}}}}}}=\color{#D61F06}{2^{3^{1^{2^{3^{2^{0^{2^{3}}}}}}}}}$ is indeed true.

In the LHS, the part after the bracket is a power of $0$ , therefore equalling $0$ , and the bracketed part will be to the power of $0$ , therefore equalling $1$ . Then we get $\boxed{2^{3}}$ .

In the RHS, there has only been a $1$ added to the chain of exponents, so the result will be the same.

This is really easy. Remember that ${a^b}^c$ is the same as $(a^b)^c$ . Remember that anything to the zero power is one, and one to the power of anything is one. Eliminating unnecessary daunting-looking bits of the equation: ${{{\cdots}^0}^2}^3 = {{{\cdots}^0}^2}^3$ ${1^2}^3 = {1^2}^3$ $1^3 = 1^3$ $1 = 1$ That looks to be true, doesn't it?

Top to bottom!!! $\begin{aligned} { 2 }^{ { 3 }^{ { { \left({ 2 }^{ { 3 }^{ 2 } }\right) }^{ { 0 }^{ { 2 }^{ 3 } } } } } } &= { 2 }^{ { 3 }^{ { { \left({ 2 }^{ { 3 }^{ 2 } }\right) }^{ { 0 } } } } } \\ &= { 2 }^{ { 3 }^{ 1 } } \\ &= { 2 }^{ { 3 } } \\ &= 8 \end{aligned}$ $\begin{aligned} { 2 }^{ { 3 }^{ { 1 }^{ { 2 }^{ { 3 }^{ { 2 }^{ { 0 }^{ { 2 }^{ 3 } } } } } } } } &= { 2 }^{ { 3 }^{ 1 } } \\ &= { 2 }^{ { 3 } } \\ &= 8 \end{aligned}$

Thus, $\displaystyle { 2 }^{ { 3 }^{ { { \left({ 2 }^{ { 3 }^{ 2 } }\right) }^{ { 0 }^{ { 2 }^{ 3 } } } } } } = { 2 }^{ { 3 }^{ { 1 }^{ { 2 }^{ { 3 }^{ { 2 }^{ { 0 }^{ { 2 }^{ 3 } } } } } } } }$ .

Therefore, the answer is True .

anything power zero is equal to 1. so 2^3=8

if we go on playing with powers.. its tough but you see that when we will proceed in this question at some point we will get LHS=2^3^(2^3^2)^0^2^3 we get 2^3^(2^3^2)^0=1(x^0=1) now we can say LHS=1^2^3..........(1) RHS=2^3^1^2^3^2^0^2^3 we get 2^3^1^2^3^2^0=1(x^0=1) now we can say RHS=1^2^3......(2) 1=2 =>LHS =RHS(hence proved

Both get simplified to 2^(3^1) which is just 2^3 making both equal... very easy to do just by looking at it with the big 0 and 1 in there which auto simplifies it.

It should have stopped at $\space \Large 2^{3^{1}}$

Once a 0 or 1 appears then there's no point of continuing

It says 2^something times 0.. So whole term becomes 2^0=1.. Same on other side..