If $a+\frac{1}{b}=1$ and $b+\frac{1}{c}=1$, then find the value of $c+\frac{1}{a}$.

Given equations:
1. $a+\frac{1}{b}=1⟹a=1-\frac{1}{b}=\frac{b-1}{b}⟹\frac{1}{a}=\frac{b}{b-1}$
2. $b+\frac{1}{c}=1⟹\frac{1}{c}=1-b⟹c=\frac{1}{1-b}$

Now, substitute $c$ and $\frac{1}{a}$ into the required expression $c+\frac{1}{a}$:

$$c+\frac{1}{a}=\frac{1}{1-b}+\frac{b}{b-1}$$

Rewrite the second fraction with a matching denominator:

$$c+\frac{1}{a}=\frac{1}{1-b}-\frac{b}{1-b}=\frac{1-b}{1-b}=1$$