An NFA with no equivalent DFA with fewer than $2^n$ states: Example and Proof
The proper way to do this proof is using Myhill&ndash;Nerode theory, which I will now explain. For a regular language $L$, say that $x,y$ are inequivalent if there exists a word $z$ such that $xz \in L$ and $yz \notin L$, or such that $xz \notin L$ and $yz \in L$. In every DFA accepting $L$, the state that is reached after reading $x$ is different from the state that is reached after reading $y$ (exercise). Hence, if we find a collection $x_1,\ldots,x_M$ of words such that $x_i,x_j$ are inequivalent for all $i \neq j$, then every DFA for $L$ must contain at least $M$ states (exercise).

For your language $L = \Sigma^* 1 \Sigma^{n-1}$, you can take as your collection of words the set of all words of length $n$. If $x,y$ are two different words of length $n$ then $x_i \neq y_i$ for some $i$; say $x_i = 0$ and $y_i = 1$. Then $x 0^{i-1} \notin L$ (since the $n$th from last bit is $x_i$ = 0), whereas $y 0^{i-1} \in L$ (since the $n$th from last bit is $y_i$ = 0). This completes the proof.