Probability of seeing sun rise tomorrow using Bayes theorem When Richard Price's published the Bayes theorem, he gave the example of a man seeing the sun-rise for the first time and wondering if it happened everyday. With each observation thereafter, he updates his belief (i.e. the sun does rise everyday) using Bayesian reasoning. This question tries to analyze the same problem.

It is a stormy night and an alien prisoner is abandoned on the frigid earth during the first ice age. Scared, he finds a cave and only comes out for a few hours at night to find some food. He has no luck though. After spending 6 nights like this, hunger becomes unbearable and so on the 7th night he decides to venture out for a little longer. After a few hours, he sees the skies changing colors. The only time he saw them change color before was during  thunderstorm lightning on the night he was dropped. This time, however, he witnesses a glorious sunrise. He came from the dark side of a tidally-locked planet (i.e.  one side always faces its star and the other  side is forever dark). On his planet, days are not tied to the sun, but time (24 hours is measured as a day). As the warmth soothes him, he wonders if this is an one off event on earth or if the sun will rise again tomorrow. How can he use Bayes theorem to update his beliefs?

Bayes Theorem
Total number of nights observed            = 7
Total number of times sky changed colors   = 2 (once due to lightning and once due to sunrise)
Total number of times the sunrise happened = 1

P(belief | observation) = P(observation | belief) x P(belief) / P(observation)

P(Belief)      = The sun will rise tomorrow
P(Observation) = Sky changed colors
P(Sun will rise | Sky changed color) = P (Sky changed color | Sun will rise) * P (Sun will rise) / P(Sky changed color)

Day 2
P(Sun will rise | Sky changed color) = (1/2) * (1/7) / (2/7)
                                 = 0.25

Day 3
P(Sun will rise | Sky changed color) = (2/3) * 0.25 / (3/8)
                                 = 0.44

Day 4
P(Sun will rise | Sky changed color) = (3/4) * 0.44 / (4/9)
                                 = 0.74

Day 5
P(Sun will rise | Sky changed color) = (4/5) * 0.74 / (5/10)
                                 = 1.184

Why does the posterior probability go beyond 1? What am I doing wrong here?

This is an extended comment rather than an answer. I don't think I understand your intention, but I'll show how I interpret the question as it's now written and how I would solve it. Maybe that will help you clarify what you're asking.
Let $A$ be the event "the sun rises", $A^c$ be "no sunrise", $B$ be "the sky changes color", and $B^c$ be "no change in color. The data observed for the first 7 days are the events (say the lightning happened on the third day)
$A^cB^c$, $A^cB^c$, $A^cB$, $A^cB^c$, $A^cB^c$, $A^cB^c$, $AB$
We now begin observing $AB$ every day, and want to find $P(A|B)$, the probability that the sun will rise given the sky changes color. I would use the definition of conditional probability, $P(A|B) = P(A,B)/P(B)$.
When we wake up on Day 8, we have 7 days of previous data and we calculate
$P(A|B) = P(A,B)/P(B) = (1/7)/(2/7) = 1/2$
On Day 9 we add the value $AB$ to the previous data
$P(A|B) = P(A,B)/P(B) = (2/8)/(3/8) = 2/3$
Day 10
$P(A|B) = P(A,B)/P(B) = (3/9)/(4/9) = 3/4$
Etc. These values converge to 1, as expected.
We could also use $P(A|B) = P(B|A)P(A)/P(B)$. Notably, in this example $P(B|A) = 1$
Day 8
$P(A|B) = P(B|A)P(A)/P(B) = \frac{(1/1)(1/7)}{(2/7)} = 1/2$
Day 9
$P(A|B) = P(B|A)P(A)/P(B) = \frac{(2/2)(2/8)}{(3/8)} = 2/3$
Day 10
$P(A|B) = P(B|A)P(A)/P(B) = \frac{(3/3)(3/9)}{(4/9)} = 3/4$

I will also remark that in your "Day 2" calculation, you use P(Sky changes color | Sun will rise) = 1/2, which is incorrect. I don't follow your calculations for Day 3 and beyond.

Further example showing updating with new independent information, similar to this youtube video from comments. Let $C$ be the event "the moon is visible", and assume that $C$ is appropriately independent of $B$ and independent of $B$ given $A$.
As the alien is exploring on the 9th day he finds a record from a previous traveler that says "The moon was visible on only 20% of the days when the sun rose" and "the moon is visible 40% of the time". That is, $P(C|A) = 0.2$, $P(C) = 0.4$. When the alien wakes up on the 10th day he observes the sky changing color, that is, he observes $B$. He calculates the probability of seeing the sun rise later that day as $3/4$ as above. Then he notices that he can see the moon, which he wasn't paying attention to before. He decides to use the traveler's information to update his belief. His prior probability of the sun rising (given he has already seen $B$) was 0.75.
$$
P(A|C) \approx P(C|A)P(A)/P(C) = \frac{0.2 * 0.75}{0.4} = 0.375
$$
This calculation was not really correct. What we were trying to calculate was $P(A|B,C)$, the probability of a sunrise given we've seen the sky change color and have seen the moon. Depending on exactly what independence assumptions between $B$ and $C$ are available the calculation we performed may be reasonable or may not be.
If we actually calculate $P(A|C)$, that would be assuming that we can see the moon but have not seen whether the sky changes color yet. On day 10 we've observed the sun rise 3 out of 9 times, so $P(A) = 3/9$. Doing that calculation gives
$$
P(A|C) = P(C|A)P(A)/P(C) = \frac{0.2*(3/9)}{0.4} = 0.167
$$
Clearly stating exactly what you're calculating and assuming is necessary to get the correct value and communicate it correctly.
