discrete mathematics - Prove that $1+a+a^2+cdots+a^n=(1-a^{n+1})/(1-a)$.

I have problem. Prove this using Mathematical Induction. I am a newbie in Mathematics. Please help me.

$$1+a+a^2+\cdots+a^n = \frac{1-a^{n+1}}{1-a}$$

This is my way for get the proof

Basic Induction:
$$ p(1)= a^1 = 1-a^1+1+1/1-a$$
$$ = 1-a^3/1-a$$

Really I don't understand this case.

Answer

The basic induction should be for n=0, then $1 = \frac{1-a}{1-a} = 1$. Now assume it's true for $n=k$ and prove it for $n=k+1$.

So, $p(k+1) = 1+a+a^2+...+a^k +a^{k+1} = \frac{1-a^{k+1}}{1-a} + a^{k+1} = \frac{1-a^{k+1}+a^{k+1} - a^{k+2}}{1-a} = \frac{1-a^{k+2}}{1-a}$ which is exactly what we want