Please explain the answer.

Example 6 prove that 
                2.7 ⁿ + 3.5 ⁿ  – 5 is divisible by 24, for all n$\in$ N.
solution Let the statement p (n) be defined as
            P(n) : 2.7 ⁿ +3.5 ⁿ  – 5 is divisible by 24. 
We note that P(n) is true for n =1, since 2.7 + 3.5  5 = 24, which is divisible by 24. 
Assume that p(k) is true 
i.e 2.7^k+3.5 ^k  – 5 = 24q, when q  $\in$  N
Now we wish to prove that P(k + 1) is true whenever P(k) is true.
We have 
            ​$$\begin{gathered} 2.7{ }^{k+1} + 3.5{ }^{k+1} – 5 = 2.7^{ k}. 7^{ l} + 3.5^k . 5^l – 5 \\ = 7 [2.7^{ k} + 3.5^{ k} – 5 – 3.5^{ k} + 5] + 3.5^{ k} . 5 – 5 \\ = 7 [24q – 3.5^{ k} + 5] + 15.5^{ k} – 5 \\ = 7 \times 24q – 21.5^{ k} + 35 + 15.5^{ k} – 5 \\ = 7 \times 24q – 6.5^{ k} + 30 \\ = 7 \times 24q – 6 (5^{ k} – 5) \\ = 7 \times 24q – 6\hspace{0.17em}\left(4p\right) \left[\right(5k – 5) \mathrm{is} \mathrm{a} \mathrm{multiple} \mathrm{of} 4 (\mathrm{why}?\left)\right] \\ = 7 \times 24q– 24p \\ = 24 (7q – p) \\ = 24 \times r, r = 7q – p, \mathrm{is} \mathrm{some} \mathrm{natural} \mathrm{number}. ...\left(2\right) \end{gathered}$$

The expression on the R.H.S. of (1) is divisible by 24. Thus P(k + 1) is true whenever P(k) true 
Hence, by principle of mathematical induction, P(n) is true for all n$\in$ N.​