Summation over a constant Let $n \geq$ 1, show by induction that: $\sum_{i=0}^{n-1} 1 = n$
Now for the base case, $n=1$, we get:  $\sum_{0}^{0} 1 = 1$, but I don't understand why this is true, we sum $0$ times, $\sum_{0}^{0} 1$, so how can this be $1$. Seems rather confusing...
Is this just by convention that for any constant $c$, we have that $\sum_{0}^{0} c = c$? Like for any $n$, $n^0 = 1$?
 A: Please note that according to the definition of capital-sigma notation, any sigma sum must be of the form$$\sum_{i=m}^na_i=a_m+a_{m+1} + \cdots + a_{n-1}+ a_n,$$where $i$ is the index of summation and $m$ and $n$ are integers.
So, why does $\sum_{i=0}^{n-1}1$ not have any indexed variable? In fact, the sum is the simplified form (by convention) of some $\sum_{i=0}^{n-1}a_i$, where for each $0 \le i \le n-1$ we have $a_i=1$ (A similar convention appears in some polynomials. For example, $x^2-1$ is the simplified form of $x^2+0x-1$; please remember that any quadratic polynomial is of the form $ax^2+bx+c$).
For example, let $a_i=1^i$. Then we have$$\sum_{i=0}^{n-1}1^i= \sum_{i=0}^{n-1}1.$$Now, for the base case of the induction proof, $n=1$, we have$$\sum_{i=0}^{0}1= \sum_{i=0}^{0}1^i=1^0=1. \, ^{\dagger }$$Let us try another example. Let $a_i=\cos 2\pi i$. Then we have$$\sum_{i=0}^{n-1}\cos 2\pi i=\sum_{i=0}^{n-1}1.$$Now, for the base case of the induction proof, $n=1$, we have$$\sum_{i=0}^{0}1= \sum_{i=0}^{0}\cos 2\pi i=\cos 2\pi (0)=1.$$

Footnote
$\dagger$ I prefer to call things like $a^0=1$ "definition" rather than "convention". In fact, raising to the power of $0$ has been defined in this way in order that the quotient rule$$\frac{a^n}{a^m}=a^{n-m}, \quad n,m \in \mathbb{N}, \quad n \gt m$$can be generalized to cover the case $m=n$ as follows.$$1=\frac{a^n}{a^n}=a^{n-n}=a^0.$$
