I want to understand the proof of
$$5^{2n}-2^n \text{ is divisble by } 23 \text{ for } n \in \mathbb{N}$$
$$\text{Base case: } n = 0 \Rightarrow5^{2\cdot 0}-2^0 = 1-1 =0 ; 0 \text{ is divisble by 23}$$
$$\text{Hypothesis: The proposition holds for }n \in \mathbb{N}, \text{ so } 5^{2n}-2^n = q \cdot 23, q \in \mathbb{N}$$
$$\text{Induction step:} 5^{2(n+1)} - 2^{n+1} = 5^{2n+2} - 2 \cdot 2^n = 25 \cdot 5^{2n}- 2 \cdot 2^n = 2 \cdot (5^{2n}-2^n) +23 \cdot 5^{2n} $$
$$= \text{(IH) } 2 \cdot 23q+23 \cdot5^{2n} = 23\cdot(2q+5^{2n}) \Rightarrow \text{ Is divisible by 23}$$
What I don't get is why $25 \cdot 5^{2n}- 2 \cdot 2^n = 2 \cdot (5^{2n}-2^n) +23 \cdot 5^{2n}= 2 \cdot 23q+23 \cdot5^{2n}$.