I'm looking for a proof verification, precisely at the end of the proof. We proceed by induction on $n$.

*Pf.* Base case $n = 1$,$$\sum\_{j=0}^1 {1 \choose j} = {1 \choose 0} + {1 \choose 1} = 1 + 1 = 2^1$$
Notice also for $n=2$, $$\sum\_{j=0}^2 {2 \choose j} = {2 \choose 0} + {2 \choose 1} + {2 \choose 2} = 1 + 2 + 1 = 2^2$$ Now, suppose that for $k\in\mathbb{N}$ it is true that $$\sum^k\_{j=0} {k \choose j} = 2^k$$
 Then, by Pascal's identity, 
$$\sum^k\_{j=0} {k \choose j} + \sum^k\_{j=0} {k \choose j-1} = \sum^{k+1}\_{j=0} {k+1 \choose j} = 2^{k+1}$$ Since $k$ is any natural number, it should then also work for $k+1$.

My reservations are that I'm using the assumption incorrectly or misusing the assumption by not starting with the assumption. Any clarity is appreciated.

Edit: It seems that changing the bound on the second sum in this equation

> 
> Then, by Pascal's identity, 
>  $$\sum^k\_{j=0} {k \choose j} + \sum^k\_{j=0} {k \choose j-1} = \sum^{k+1}\_{j=0} {k+1 \choose j} = 2^{k+1}$$
> 
> 
> 

to $k+1$ yielded the quickest fix to the original proof, with the convention that, as Brian M. Scott pointed out: ${n \choose k}$ is 0 if $k$ is a negative integer.