Counting argument proof or inductive proof of $F_1 {n \choose1}+...+F_n {n \choose n} = F_{2n}$ where $F_i$ are Fibonacci

(The following answer is essentially Example 3.1.5 in this pdf.)

First, $F_{2n}$ counts the number of ways of tiling a strip of length $2n-1$ with tiles of length either $1$ (squares) or $2$ (dominos).

Now, let's look at the number of tilings of that strip in which $k$ of the first $n$ tiles are squares. There are $\binom{n}{k}$ ways of arranging the first $n$ tiles. A collection of $k$ squares and $n-k$ dominos will have length $2n-k$. So the strip remaining after the first $n$ tiles has length $k-1$, which means that it can be tiled in $F_k$ ways. In summary, the number of tilings of this sort is $F_k \binom{n}{k}$.

Moreover any tiling of the strip uses at least $n$ tiles (or it would have length at most $2n-2$), and the first $n$ tiles cannot all be dominos (or they would have length $2n$). So we have $$ F_{2n}=\sum_{k=1}^n F_n \binom{n}{k} $$ as desired.

This is one way to attack the problem by induction. We need to generalize the statement a little bit before we can carry out the induction step.

Extend definition of $F_n$ to $F_0 = 0$. For any $n > 0$, let $\mathcal{S}_n$ be the statement:

$$\mathcal{S}_n :\quad F_{2n+k} = \sum_{\ell=0}^n \binom{n}{\ell} F_{\ell+k},\;\text{ for all k} \ge 0$$ When $n = 1$, $S_n$ reduces to $F_{k+2} = F_{k+1} + F_k$ for all $k \ge 0$. This is trivially true.

Assume $\mathcal{S}_{n}$ is true. For any $k \ge 0$, we have

$$\begin{align}F_{2(n+1)+k} = F_{2n+(k+2)} &\stackrel{\mathcal{S}_n}{=} \sum_{\ell=0}^n \binom{n}{\ell} F_{\ell+(k+2)} = \sum_{\ell=0}^n \binom{n}{\ell} \left( F_{\ell+k+1} + F_{\ell+k}\right)\\ &= \sum_{\ell=1}^{n+1}\binom{n}{\ell-1}F_{\ell+k} + \sum_{\ell=0}^n \binom{n}{\ell} F_{\ell+k}\\ &= F_{n+1+k} + F_k + \sum_{\ell=1}^n \left(\binom{n}{\ell-1}+\binom{n}{\ell}\right)F_{\ell+k} \end{align} $$ Notice $\binom{n+1}{\ell} = \binom{n}{\ell}+\binom{n}{\ell-1}$ for $1 \le \ell \le n$, this leads to $$F_{2(n+1)+k} = F_{n+1+k} + F_{k} + \sum_{\ell=1}^n \binom{n+1}{\ell}F_{\ell+k} = \sum_{\ell=0}^{n+1} \binom{n+1}{\ell}F_{\ell+k}$$ Since this is valid for all $k$, $\mathcal{S}_{n+1}$ is true. By principle of induction $\mathcal{S}_n$ is true for all $n$.
In particular, if we fix $k$ to $0$, the desired identity follows: $$F_{2n} = \sum_{\ell=0}^n \binom{n}{\ell}F_{\ell} = \sum_{\ell=1}^n \binom{n}{\ell}F_{\ell}$$