Don't Use a Calculator!

You can simplify both expressions:

${ 30 }^{ 40 }{ \qquad 45 }^{ 30 }$

Take the tenth root of both sides

$\sqrt [ 10 ]{ { 30 }^{ 40 } } \qquad \sqrt [ 10 ]{ { 45 }^{ 30 } }$

${ 30 }^{ 4 }{ \qquad 45 }^{ 3 }$

This equals

${ (15\times 2) }^{ 4 }{ \qquad (15\times 3) }^{ 3 }$

Divide out the cube of 15

$\frac { { (15\times 2) }^{ 4 } }{ { 15 }^{ 3 } } \qquad \frac { { (15\times 3) }^{ 3 } }{ { 15 }^{ 3 } }$

$15\times { 2 }^{ 4 }\qquad { 3 }^{ 3 }$

$15\times 16\qquad 27$

$15\times 16$ is clearly greater than $27$