Playing roulette using martingale

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A player with unlimited money decides to play roulette. He bets $1$ on red, if he loses, he bets $2$, if he loses again he bets $4$ and so on till he wins. Prove that he is guaranteed to make a profit.

I know this is the theory of martingale and I looked for proof online but everything I found was way more complex than what's needed here. Isn't there a formula to prove that after unlimited tries the probability of never winning is $0$?

I understand why it works but I'm having a hard time explaining it using mathematics. Any help would be greatly appreciated.

Thank you

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Let's assume for simplicity that the probability $p$ of winning a given spin is $$p=0.5$$ and then the probability of losing $q$ on a given hand is $$q=1-p=0.5$$

With the martingale betting system, one win will yield a net profit because you always bet more than you've lost. The probability you will not win at least once after $n$ spins is $$q^{n}=0.5^n$$

The limit of this expression as $n\to\infty$ then represents the probability that you will never win. The limit is $$\lim_{n\to\infty}0.5^n = 0$$

Therefore the roulette player will almost surely win at least once, yielding a net profit.