Combinatorial proof of $\sum_{j=0}^{k} \binom{n}{j} = \sum_{j=0}^k \binom{n-1-j}{k-j}2^j$

Give a combinatorial proof for this identity for non-negative integer $k$ and $n$ such that $0 \leq k < n$ $$\sum_{j=0}^{k} \binom{n}{j} = \sum_{j=0}^k \binom{n-1-j}{k-j}2^j$$

My attempt: I tried to reduce this identity. Since $k,n$ are nonnegative integer and $k<n$ then I can change the upper limit into

$$\sum_{j=0}^{n-1} \binom{n}{j} =2^n-1 \quad \text{and} \quad \sum_{j=0}^{n-1} \binom{n-1-j}{n-1-j}2^j = \sum_{j=0}^{n-1} 2^j ~.$$

So, I just need to show $$\sum_{j=0}^{n-1} 2^j =2^n-1 $$ using combinatorial proof. Am I right? I really need helps

Big hint

$\sum_{j=0}^k\binom{n}j$ counts subsets of $\{1,2,\dots,n\}$ with at most $k$ elements. Such a subset is missing at least $n-k$ elements. Arrange the missing elements in a sorted list, so the first element in the list is the smallest missing element. For how many such subsets is the $(n-k)^{th}$ entry of the list of missing elements equal to $n-j$?

Further explanation, just shy of a full solution:

If a subset satisfies this condition, then the element $n-j$ is missing, and among the elements $\{1,2,\dots,n-j-1\}$, exactly $n-k-1$ are missing. The elements above $n-j$ are either included or excluded, arbitrarily.

Here is an illustration for $n=5,k=3$. We represent subsets as a string of zeroes and ones. The $(n-k)^{th}$ smallest, i.e. second smallest, missing element is highlighted in red. $$ \begin{array}{c|c|c|c} \binom{5-1-0}{3-0} & \binom{5-1-1}{3-1}2 & \binom{5-1-2}{3-2}2^2 & \binom{5-1-3}{3-3}2^3 \\\hline 1110\color{red}0 & 110\color{red}00 &10\color{red}000 &0\color{red}0000 \\ 1101\color{red}0 & 110\color{red}01 &10\color{red}001&0\color{red}0001 \\ 1011\color{red}0 & 101\color{red}00 &10\color{red}010&0\color{red}0010 \\ 0111\color{red}0 & 101\color{red}01 &10\color{red}011&0\color{red}0011 \\ & 011\color{red}00 &01\color{red}000&0\color{red}0100 \\ & 011\color{red}01&01\color{red}001&0\color{red}0101 \\ & &01\color{red}010&0\color{red}0110 \\ & &01\color{red}011&0\color{red}0111 \end{array} $$