An heuristic argument why factoring is difficult

Let $n=pq$ be the prime decomposition of $n$ and let $n=2^e$. Let us suppose that $p = q = \sqrt{n}$. Let $p_n$ be the probability to have a number $1 \le x \le n$ with $1 < gcd(x,n) < n$ and as such to also have a non-trivial factor of $n$. Then we have $$p_n = 1-\frac{\phi(n)}{n}=1-\prod_{p|n}{(1-\frac{1}{p})} = 1-(1-\frac{1}{p})(1-\frac{1}{q})$$ $$= 1-(1-\frac{1}{\sqrt{n}})^2 = 1-(1-\frac{1}{\sqrt{2^e}})^2$$ Suppose there exists a factoring-algorithm which runs in polynomial time $f(e)$ and which draws a number $x$ at every step and computes $gcd(x,n)$. Let $X$ be the number of numbers $x$ with $1 \lt x \lt n$ and $1 \lt gcd(x,n) \lt n$ which the algorithm finds after $f(e)$ steps. Then by definition of the algorithm we must have $$1 = P(X \ge 1)$$ But on the other hand we have $$P(X\ge 1) = 1 - P(X=0) = 1 - (1-p_n)^{f(e)}$$ $$= 1-(1-\frac{1}{\sqrt{2^e}})^{2 \cdot f(e)}$$ The last equality is by definition of the algorithm valid for every $e$. But for $e \rightarrow \infty$ we have $$1 = P(X \ge 1) = 1-(1-\frac{1}{\sqrt{2^e}})^{2 \cdot f(e)} \rightarrow_{e \rightarrow \infty} 0$$ hence for $e \rightarrow \infty$ we have the contradiction $1 = 0$.