1. **State the problem:** Prove that the sum of binomial coefficients from 0 to n equals $2^n$, i.e., $$\sum_{r=0}^n {n \choose r} = 2^n.$$ 2. **Base case (n=1):** Calculate $$ {1 \choose 0} + {1 \choose 1} = 1 + 1 = 2 $$ which equals $$2^1 = 2$$. So, the statement is true for n=1. 3. **Inductive hypothesis:** Assume it is true for $$n=k$$, i.e., $$\sum_{r=0}^k {k \choose r} = 2^k.$$ 4. **Inductive step:** We need to prove the statement for $$n=k+1$$: $$\sum_{r=0}^{k+1} {k+1 \choose r} = ?$$ 5. **Use Pascal's rule:** Recall that $$ {k+1 \choose r} = {k \choose r} + {k \choose r-1} $$ for $$1 \leq r \leq k$$. 6. **Expand the sum:** $$ \sum_{r=0}^{k+1} {k+1 \choose r} = {k+1 \choose 0} + \sum_{r=1}^k {k+1 \choose r} + {k+1 \choose k+1} $$ With Pascal's identity: $$ = 1 + \sum_{r=1}^k \left({k \choose r} + {k \choose r-1}\right) + 1$$ 7. **Rewrite the sums:** Separate the sums: $$ = 1 + \left(\sum_{r=1}^k {k \choose r}\right) + \left(\sum_{r=1}^k {k \choose r-1}\right) + 1 $$ Shifting index in second sum: $$ \sum_{r=1}^k {k \choose r-1} = \sum_{s=0}^{k-1} {k \choose s} $$ 8. **Evaluate sums:** $$ \sum_{r=1}^k {k \choose r} = \sum_{r=0}^k {k \choose r} - {k \choose 0} = 2^k - 1 $$ $$ \sum_{s=0}^{k-1} {k \choose s} = \sum_{r=0}^k {k \choose r} - {k \choose k} = 2^k - 1 $$ 9. **Combine and simplify:** $$ 1 + (2^k - 1) + (2^k - 1) + 1 = 2^k + 2^k = 2 imes 2^k = 2^{k+1} $$ 10. **Conclusion:** Therefore, $$\sum_{r=0}^{k+1} {k+1 \choose r} = 2^{k+1},$$ and by induction, the formula holds for every positive integer $$n$$. **Final answer:** $$\sum_{r=0}^n {n \choose r} = 2^n.$$