Use the definition to prove that $\sum_{j=0}^n \binom{n}{j}=2^n $

Question

This is a question from Cambridge, Downing sample test

Using the definition $\left( \begin{array}{c} \text{n}\\ \text{k}\\ \end{array} \right) =\frac{\text{n!}}{\text{k!}\left( \text{n}-\text{k} \right) !} $ , prove that $ \sum_{\text{j}=0}^{\text{n}}{\left( \begin{array}{c} \text{n}\\ \text{j}\\ \end{array} \right) =2^{\text{n}}} $

Of course I know how to prove this by considering the subsets, or by binomial expansion.

But how to do it by definition? No idea about it.

Answer 1

Let $[x^m]f(x)$ stand for the coefficient of $x^m$ in the Maclaurin series of $f(x)$. We have: $$\sum_{k=0}^{n}\frac{n!}{k!(n-k)!}=n!\sum_{k=0}^{n}\frac{1}{k!}\cdot\frac{1}{(n-k)!}=n!\cdot[x^n]\left(\sum_{k\geq 0}\frac{x^k}{k!}\right)^2=n!\cdot[x^n]e^{2x}=n!\cdot\frac{2^n}{n!}=2^n. $$