Gambling game: What is the probability of eventually going broke

Let $p(n)$ be the probability of going broke starting with $n$ dollars. On the one hand, we have $$ p(n)=p(1)^n\tag1 $$ because moving from $\$n$ to $\$0$ is the same as moving from $\$k$ to $\$k-1$ a total of $n$ times, for $k=n,n-1,n-2,\dots,1$. On the other hand, we must have $$ p(1)=\frac34 p(2)+\frac14\qquad (n\ge1)\tag2 $$ Substituting $(1)$ into $(2)$, and setting $n=1$, you get $p(1)=\frac34 p(1)^{2}+\frac14$, the solution to which is $p(1)=1/3$ or $p(1)=1$. If we had $p(1)=1$, then we would have $p(n)=1$ for all $n$, so that you certainly go broke no matter how much money you start with, which is clearly incorrect.$^*$ Therefore, $$ \bbox[3pt,border: 1.5pt black solid]{p(n)=(1/3)^n.} $$ $^*$A rigorous proof of this certainly exists, perhaps by applying Stirling's approximation to an exact combinatorial expression for $p(n)$, or with some central limit theorem argument.

Edit: For a more convincing argument, suppose that the process does not stop when you hit zero. If $p(1)=1$, you would certainly hit $\$0$, and from there certainly hit $\$(-1)$, and from there $\$(-2)$, and so on. This means you would drift arbitrarily far down. This contradicts the law of large numbers, which says $S_n/n$ converges to $1/2$ almost surely as $n\to\infty$, where $S_n$ is your winnings after $n$ plays.

As another possible method. Consider a slightly modified game where the game can end if the money reaches $0$ or $N$ dollars (we can take the limit later). Let $P_i$ be the probability that you win starting with $i$ dollars. Suppose the probability of winning a single game is $p$ and $q = 1- p$ is the probability of losing a single game. We can observe the relation $$ P_i = p P_{i+1} + q P_{i-1}. \tag{1} \label{1}$$ This is a linear difference equation which can easily be solved for roots (via $P_i = z^i$). The result is $$P_i = \begin{cases} A + B \left(\frac{q}{p} \right)^i & \text{if } p \neq q, \\ A + B i & \text{if } p = q = 1/2. \end{cases}$$ Then noting the boundary conditions $P_0 = 0$ and $P_N = 1$, we obtain, $$P_i = \begin{cases} \frac{1-\left(\frac{q}{p} \right)^i}{1- \left(\frac{q}{p} \right)^N} & \text{if } p \neq q, \\ \frac{i}{N} & \text{if } p = q = 1/2. \end{cases}$$ Taking the limit as $N \to \infty$, yields the desired result.