What is uniformity in Boolean circuits exactly? I have two questions on Kaveh's answer to Definition of uniform boolean circuit :

Kaveh mentions that the input is in unary encoding. In the definition it says the input is $1^n$, afaik $1^n$ is  a sequence of 1's repeated n times, but unary encoding is a sequence of 1's  and ends with 0 there is no power of 1 that gives 0... So how is $1^n$ a unary encoding?
Why do we use a unary encoding and not the binary encoding of n? what happens if we use binary instead?
Here is the binary encoding of 5: $$101$$ And here is its unary encoding: $$11111$$ Hopefully the difference is apparent.
Let's see what happens if we switch to binary encoding in the context of the linked question. Given the binary encoding of $n$, we need to produce a circuit, operating on $n$ bits, in logspace, and so in polynomial time. How much time do we have? The binary encoding of $n$ has length roughly $\log n$, so we have time $O(\log^C n)$, which isn't enough to produce a circuit which reads all input bits.
In contrast, if we use unary encoding, then the length of the input is $n$, and so we have time $O(n^C)$, which is more reasonable (in addition, we are only allowed to use $O(\log n)$ space).