Sum of combinations of the n by consecutive k

In a book, I found that the sum of combinations: $\binom{n}{k} + \binom{n}{k+1} +\cdots+ \binom{n}{n}$, where k starts from 0, equals $2^n$. It is possible to express this statement via sum: $2 + \sum_{k=1}^{n-1}{n!\over k!(n-k)!} = 2^n$, using binomial theorem. I tried several small n values, such as 4, 6 and others, the statement looks correct. But I can’t find a regularity, respectively I can’t understand, how to prove the truth of the statement?

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