Memoryless property and geometric distribution

We have $$P(X\geq s+t\mid X>t)=P(X\geq s)$$ and $$P(X\geq s+t\mid X>t)=\dfrac{P(X\geq s+t)}{P(X>t)}$$ therefore$$P(X\geq s+t)=P(X\geq s)P(X>t).$$ Summing the last equality once over $s$ then once over $t$ we obtain $$I: \sum_{s=1}^{\infty}P(X\geq s+t)=\sum_{s=1}^{\infty}P(X\geq s)P(X>t)$$ $$II: \sum_{t=1}^{\infty}P(X\geq s+t)=\sum_{t=1}^{\infty}P(X\geq s)P(X>t)$$ Now in $I$ set $t=k$, and set $s=k$ in $II$ so that their left hand side be equal and consequently $$P(X\geq k)\sum_{t=2}^{\infty}P(X\geq t)=P(X\geq k+1)\sum_{t=1}^{\infty}P(X\geq t).$$ Now let $\alpha=\sum_{t=2}^{\infty}P(X\geq t)$ and evaluate for $k=1$ to obtain, $$\alpha=\frac{1-p}{p},$$where $p=P(X=1).$ Therefore $(1-p)P(X\geq k)=P(X\geq k+1)$. Since $P(X\geq 1)=1$ we have that $$P(X\geq k)=(1-p)^{k-1}; k=1,2,...$$ implying that $$P(X=k)=(1-p)^{k-1}-(1-p)^{k}=p(1-p)^{k-1}.$$Therefore $X\sim$ Geometric$(p)$.