Probability that n is prime such that n fails the Miller-Rabin test N times I'm working through An Introduction to Mathematical Cryptography and one of the exercises asks


  Suppose that we run the Miller–Rabin test N times on the integer n and that it fails to prove that n is composite. Show that the probability that n is prime satisfies (approximately)
  
  Pr($n$ is prime $\mid$ the Miller–Rabin test fails $N$ times) $\geq 1 − \frac{ln(n)}{4^N}$.


This is assuming that the Miller-Rabin test is 75% accurate and that the probability $n$ is prime is $\frac{1}{ln(n)}$. I think by "failing on a number", they mean they identify the number as prime. (Fail to identify it as composite.)

Based on the assumptions, I find the following:

Let $n$ be "$n$ is prime" and $N$ be "The Miller-Rabin test fails $N$ times". Because the test always fails on a prime, $P(N \mid n)=1$. $P(n)=\frac{1}{ln(n)}$ as above, and $P(N \mid n^c)=(\frac{1}{4})^N$ because the test identifies a composite as prime $\frac{1}{4}$ of the time.

From this, I can use the fact that $P(n \mid N)=\frac{P(N \mid n)P(n)}{P(N \mid n)P(n) + P(N \mid n^c)P(n^c)}$ to find

$$
P(n \mid N)=\frac{\frac{1}{ln(n)}}{\frac{1}{ln(n)} + (\frac{1}{4^N})(1-\frac{1}{ln(n)})}\\
=\frac{\frac{1}{ln(n)}}{\frac{1}{ln(n)} + \frac{1}{4^N}-\frac{1}{4^Nln(n)}}\\
=\frac{\frac{4^N}{ln(n)}}{\frac{4^N}{ln(n)} + 1-\frac{1}{ln(n)}}\\
=\frac{4^N}{4^N + ln(n)-1}\\
$$

But that doesn't seem to get me anywhere. What am I missing here?
Just one step more, your last expression is

$$\frac{4^N}{4^N +\ln(n)-1}=1-\frac{\ln n -1}{4^N+\ln n -1}\geq 1-\frac{\ln n -1}{4^N}\geq 1-\frac{\ln n}{4^N}.$$