There is a cute argument for $b=c+1$ though I don't know how to pull it through in the general case.
Let $U$ be the set of ordered $c+1$-tuples of sets of cardinality $c$ in which no set repeats. Then, obviously, $|U|=(c+1)!{{n\choose c}\choose c+1}$. Let $V$ be the set of ordered $c$-tuples of sets of cardinality $c+1$ in which no set repeats. Then $|V|=c!{{n\choose c+1}\choose c}$. Now take any element of $V$, i.e., an ordered family of sets $A_1,\dots,A_{c}$. Take any $a_1\in A_1$ ($c+1$ choices). Suppose that $a_1\in A_1,\dots,a_k\in A_k$ are already chosen. Then we want to choose $a_{k+1}\in A_{k+1}$ so that $a_{k+1}\ne a_j$ and $A_{k+1}\setminus \{a_{k+1}\}\ne A_j\setminus\{a_j\}$ for all $j=1,\dots,k$. Suppose that we have $\ell$ prohibitions of the first type. Then, if we honor them, the prohibitions of the second type for the corresponding sets will be honored automatically (if $a_j\in A_{k+1}$ and we don't remove it, then surely $A_{k+1}\setminus \{a_{k+1}\}\ne A_j\setminus\{a_j\}$. The remaining $k-\ell$ sets introduce at most one prohibition of the second type each, so we have $\le k$ prohibitions total. Thus we have $\ge c+1-k$ choices for $a_{k+1}$. Once we run the whole procedure, the reduced sets $A_j$ listed in the same order and the set $\{a_1,\dots,a_c\}$ listed last will constitute an element of $U$ and each element of $V$ will generate $\ge(c+1)!$ different elements of $U$ (by different choices of $a_1,\dots,a_c$. On the other hand, each element of $U$ can be obtained from at most $c!$ elements of $V$ (that is the number of ways to distribute the elements of the last $c+1$-st set among the first $c$. This immediately gives a non-strict inequality you want. To make it strict, you need to assume $c>1$ and $n\ge c+1$ (otherwise you have equality), in which case the ordered sequence of sets $A_j=\{1,2,\dots,c+1\}\setminus\{j\}$ has fewer than $c!$ distribution options (none at all, really) resulting in a legitimate element of $V$.
Edit (the general case)
In this case it will be convenient to define $U$ as the set of all ordered $b$-tuples of sets of cardinality $c$ such that the sets are pairwise different (as sets), the first $c$ sets are unordered, and the last $b-c$ sets are ordered. Then $|U|=b!(c!)^{b-c}{{n\choose c}\choose b}$. Let $V$ be the same as before (ordered $c$-tuples of unordered subsets of cardinality $b$ in which the sets are pairwise different), so $|V|=c!{{n\choose b}\choose c}$.
Now let $\langle A_1,\dots, A_c\rangle$ be an element of $V$. We want to remove $b-c$ elements $a_{j,1},a_{j,2},\dots,a_{j,b-c}$ from $A_j$ and form $b-c$ new ordered sets $B_k=\langle a_{1,k},a_{2,k},\dots, a_{c,k}\rangle$ ($k=1,\dots,b-c$) to get an element of $U$.
To this end start with choosing $a_{1,1},\dots,a_{1,b-c}\in A_1$ in an arbitrary way, which gives $b(b-1)\dots(b-c+1)$ choices. Now, when choosing $a_{k+1,1}\in A_{k+1}$ for $k\ge 1$, add to the standard prohibitions $a_{k+1,1}\ne a_{j,1}$, $A_{k+1}\setminus\{a_{k+1,1}\}\ne A_j\setminus\{a_{j,1}\}$ ($j\le k$), which exclude at most $k$ elements just as before, the prohibitions $a_{k+1,1}\ne a_{1,m}$ ($m>1$), which exclude additional $b-c-1$ elements at most, leaving $c+1-k$ choices as before. These additional prohibitions guarantee that the set $B_1=\{a_{1,1},a_{2,1},\dots, a_{c,1}\}$ does not contain any of the elements $a_{1,m}$ ($m>1$) and, therefore, will be different from every set $B_m$ ($m>1$) as an unordered set and we still have $c!$ choices.
Having constructed $B_1$, we construct $B_2$ with the standard restrictions $a_{k+1,2}\ne a_{j,2}$, $A_{k+1}\setminus\{a_{k+1,1},a_{k+1,2}\}\ne A_j\setminus\{a_{j,1},a_{j,2}\}$ ($j\le k$) and additional restrictions $a_{k+1,2}\ne a_{1,m}$ but now with $m>2$. This gives $c!$ choices again, and so on. Thus, from each element of $V$, we get $b(b-1)\dots(b-c+1) (c!)^{b-c}$ distinct elements of $U$. The recovery of an element of $V$ from an element of $U$ is now possible in just one way, if at all, which, again, gives a non-strict inequality. I leave it to you to figure out when and how it becomes strict in this case.
Best Answer
I will first try to explain why the result is true in the $i=0$ case. Write it as $$ 1 = \sum_{j=1}^{n} (-1)^{j-1} \binom{n}{n-j} \binom{n+j-1}{j} $$ which is $$ 0 = \sum_{j=0}^{n} (-1)^{j} \binom{n}{n-j} \binom{n+j-1}{j}. $$ Then suppose you have to put $n$ balls in $n$ boxes. For $j$ = $0$ to $n$, let there be $j$ red balls and $n-j$ blue balls. The first binomial factor in each term $\displaystyle\binom{n}{n-j}$ is the number of ways of putting one blue ball in each of $n-j$ boxes, and the second, $\displaystyle \binom{n+j-1}{j}$, is the number of ways of putting $j$ red balls in the $n$ boxes, with as many in each box as you like.
Now consider an arrangement of the balls in the boxes with exactly $k$ of the boxes used. This can only happen in the cases where $n-j$ is at most $k$, so suppose $j=n-k+r$, where $r=0\ldots k$. Then we count this arrangement $\displaystyle \binom{k}{r}$ ways (the number of ways of choosing which of the $k$ boxes doesn't have a blue ball in). But now, in the whole sum, we count how many times this arrangement appears: this is the alternating sum of a whole row of Pascal's triangle, so zero.
An algebraic version of the argument above is as follows. $$ \binom{n}{n-r} \binom{n-1}{r} \binom{n-r}{n-j} = \binom{n}{n-j} \binom{j}{r} \binom{n-1}{r} $$ (just by rearranging the factorials), so $$ \sum_{r=0}^j\binom{n}{n-r} \binom{n-1}{r} \binom{n-r}{n-j} = \binom{n}{n-j}\sum_{r=0}^j \binom{j}{j-r} \binom{n-1}{r}=\binom{n}{n-j} \binom{n+j-1}{j} $$ Now insert this in the sum above. $$ \sum_{j=0}^{n} (-1)^{j} \binom{n}{n-j} \binom{n+j-1}{j}=\sum_{j=0}^{n} (-1)^{j}\sum_{r=0}^j\binom{n}{n-r} \binom{n-1}{r} \binom{n-r}{n-j} $$ Interchanging the order of the sums, gives $$ \sum_{r=0}^{n}\binom{n}{n-r} \binom{n-1}{r} \sum_{j=r}^n (-1)^{j}\binom{n-r}{n-j}=0 $$ because the sum over $j$ is zero.
Now all you have to do is extend to $i \neq 0$!