Part 1

Recall the law of indices :

(ab)ⁿ = aⁿbⁿ

So for the first term in the third step,


(2x)ⁿ = 2ⁿxⁿ

For the second term, using the same law for the term inside the brackets,

n(2x)ⁿ⁻¹

= n × (2ⁿ⁻¹ xⁿ⁻¹)

= n × 2ⁿ⁻¹ × xⁿ⁻¹

= 2ⁿ⁻¹ × n × xⁿ⁻¹
(just the commutative law of multiplication)

= 2ⁿ⁻¹ n xⁿ⁻¹

For the third term :

n(n-1)/2! (2x)ⁿ⁻²

= n(n-1)/2 × (2x)ⁿ⁻²

[Since 2! = 2 x 1 = 2]

= n(n-1)/2 × 2ⁿ⁻² × xⁿ⁻²

(Using the indices law for the brackets as seen in the previous terms)

= n(n-1) × 2⁻¹ × 2ⁿ⁻² × xⁿ⁻²

[Recall the negative indices property a⁻¹ = 1/a
So (n-1)/2 = n(n-1) × 1/2 = n(n-1) × 2⁻¹ ]


= n(n-1) × 2⁻¹⁺ⁿ⁻² × xⁿ⁻²

[Recall the indices law aⁿ × aᵐ = aⁿ⁺ᵐ ]

= n(n-1) × 2ⁿ⁻³ xⁿ⁻²

= 2ⁿ⁻³ n(n-1) xⁿ⁻²

For the fourth term, 3! = 3 × 2 × 1 = 3 × 2

So,

n(n-1)(n-2) / 3!

= n(n-1)(n-2) / (3×2)


The idea is to keep the 3 in the denominator and change the 2 into a term with negative indices.

= n(n-1)(n-2) / 3 × 1/2

= n(n-1)(n-2) / 3 × 2-¹

From here you can combine the 2-¹ with the 2ⁿ-³ in the brackets (after you apply the indices law) to get 2ⁿ-⁴

The rest of the steps are just rearrangement.