How would I go about writing this proof in a formal way? Let c ∈ Z: Write a detailed structured proof to prove the statement:
If c^5 + 7 is even, then c is odd.
I started out like this:
Assume c ∈ Z
    Assume c^5 + 7 == 2n
        Then c == 2n + 1

Also, is this claim true? I plugged in odd numbers for c, and haven't encountered a counter-example. Is there a way to determine the veracity of the claim before doing the proof?
 A: You have left out quite a bit in between the "Assume" and "Then".
That said, might I suggest using the Contrapositive?
It's logically equivalent , and states:

If $c$ is even, then c^5 + 7 is odd.

This is quite simple to prove. Suppose that $c = 2k$ for some integer $k$. Then $c^5 + 7 = (2k)^5 + 7 = 2(16k^5 + 3) + 1$, which is odd. QED.
A: We have these equivalences
$$c^5+7\ \text{is even}\iff c^5\ \text{is odd}\iff c \ \text{is odd}$$
The last equivalence can be proved using a contrapositive proof and a direct proof
$\Rightarrow)$ by contrapositive: If $c=2n$ then $c^5=2\times 2^4n^5$ is even.
$\Leftarrow)$ If $c=2n+1$ then $(2n+1)^5=\sum_{k=0}^5{5\choose k}(2n)^k=1+\underbrace{\sum_{k=1}^5{5\choose k}(2n)^k}_{even}$
A: Your conclusion needs to be 'Then $c = 2m + 1$' for some other integer $m$ (it probably won't be the case that $c^5 + 7 = 2n$ and $c = 2n + 1$ for the same $n$).
Your first two steps are fine, though I would maybe expand the second step and say:


*

*Assume $c\in\mathbb{Z}$

*Assume $c^5 + 7$ is even

*Then $c^5 + 7 = 2n$ for some $n\in\mathbb{Z}$


The next steps could go something along the lines of proving the statements


*

*If $c^5 + 7$ is even, then $c^5$ is odd

*If $c^5$ is odd, then $c$ must be odd.


A totally different approach
You could also try arguing the contrapositive, which is
'if $c$ is not odd, then $c^5 + 7$ is not even'
i.e.
'if $c$ is even, then $c^5 + 7$ is odd'
A: The claim is indeed true.
Now, how to proceed: if $c\in\mathbb Z$ then $c^5$ has the same parity than $c$, let's see how to write the proof later.
Therefor if $c^5$ is odd, then $c$ is odd.  And $c^5$ is odd because $c^5+\text{odd}=\text{even}$.
Now you have to write it backwards:


*

*If $c^5+7$ is even, then $c^5$ is odd (because $7$ is odd and $\text{even}-\text{odd}$ is odd).

*If $c^5$ is odd for an integer $c$, then $c$ is odd.  (by contradiction: if $c$ were even, then there is $k\in\mathbb Z$ so that $c=2k$ and $c^5=32k^5=2(16k^5)$ which is even.)

*Therefor $c$ is odd.

