Extracting term from sum with index changes, where do I is misunderstood the solution?

Question

It is proper solution for $ P(X > x) $ for Geometric distribution; $$ P(X > x) = p \sum_{k=x+1}^{\infty} (1-p)^k = p (1-p)^x \sum_{k=0}^{\infty} (1-p)^k = \frac{p (1-p)^x}{1-(1-p)} = (1-p)^x $$ I am wondering how to achieve $ k = 0 $ in sum. For me it looks that the $ k $ should equal 1, like this: $$ p \sum_{k=x+1}^{\infty} (1-p)^k = p (1-p)^x \sum_{k=1}^{\infty} (1-p)^k $$ Because $$ p \sum_{k=x+1}^{\infty} (1-p)^k = p ((1-p)^{x+1} + (1-p)^{x+2} + \dots + (1-p)^{x+n} + \dots) = p (1-p)^x ((1-p)^{1} + (1-p)^{2} + \dots + (1-p)^{n} + \dots) = p (1-p)^x \sum_{k=1}^{\infty} (1-p)^k $$ Can somebody help me and tell what I am missing? I know that my solution is wrong.

Answer 1

You are right. $k$ ranges from $1$ to $\infty$. The answer is $P(X>x) =(1-p)^{x+1}$.