Proving $\sum_{t=1}^j \sum_{r=1}^{t} (-1)^{r+t} \binom{j-1}{t-1} \binom{t-1}{r-1} f(r) = f(j)$ I've run into the following identity and trying to prove it:

Let $j \in \mathbb{N}$ and $f:\mathbb{N} \to \mathbb{R}$, then
  $$
\sum_{t=1}^j \sum_{r=1}^{t} (-1)^{r+t} \binom{j-1}{t-1} \binom{t-1}{r-1} f(r) = f(j)
$$

I've so far tried to find some connection with the multinomial theorem, since the product of the two binomials in the expression is just multinomial $\binom{j-1}{k1,k2,k3}$. So perhaps something like $(f(j)-1+1)^{j-1}$, but it does not quite fit. I am missing something, any ideas how to proceed?
Edit: Further attempt, trying to collect "coefficients" of $f(r)$ yields 
$$
\sum_{t=r}^j(-1)^{r+t}\binom{j-1}{t-1}\binom{t-1}{r-1} = \sum_{t=r}^j (-1)^{r+t}\binom{j-1}{t-1} 
$$
It should be enough to show that this is $1$ when $j=r$ and $0$ otherwise. First one is simple
$$
\sum_{t=r}^r (-1)^{r+t}\binom{r-1}{t-1}  = (-1)^{2r} \binom{r-1}{r-1} = 1
$$ but how to show it is equal to $0$ in other cases ($j\neq r $) ...
 A: We seek to verify that
$$\sum_{k=1}^n \sum_{q=1}^k (-1)^{k+q}
{n-1\choose k-1} {k-1\choose q-1} f(q) = f(n).$$
Exchange sums to obtain
$$\sum_{q=1}^n \sum_{k=q}^n (-1)^{k+q}
{n-1\choose k-1} {k-1\choose q-1} f(q)
\\ = \sum_{q=1}^n f(q) \sum_{k=q}^n (-1)^{k+q}
{n-1\choose k-1} {k-1\choose q-1}.$$
Now we get for the inner coefficient
$${n-1\choose k-1} {k-1\choose q-1}
= \frac{(n-1)!}{(n-k)! (q-1)! (k-q)!}
= {n-1\choose q-1} {n-q\choose n-k}.$$
This yields for the sum
$$\sum_{q=1}^n f(q) {n-1\choose q-1} (-1)^q
\sum_{k=q}^n {n-q\choose n-k} (-1)^k
\\ = \sum_{q=1}^n f(q) {n-1\choose q-1}
\sum_{k=0}^{n-q} {n-q\choose n-q-k} (-1)^k
\\ = \sum_{q=1}^n f(q) {n-1\choose q-1}
\sum_{k=0}^{n-q} {n-q\choose k} (-1)^k.$$
We get 
$$\sum_{q=1}^n f(q) {n-1\choose q-1} [[n-q = 0]]
= f(n) {n-1\choose n-1} = f(n).$$
A: $$
\begin{align}
&\sum_{t=1}^j\sum_{r=1}^t(-1)^{r+t}\binom{j-1}{t-1}\binom{t-1}{r-1}\,f(r)\\[3pt]
&=\sum_{r=1}^j\underbrace{\sum_{t=r}^j(-1)^{t-r}\binom{j-1}{r-1}\binom{j-r}{t-r}}\,\,f(r)\tag{1}\\
&=\sum_{r=1}^j\hspace{14mm}\underbrace{0^{\,j-r}\binom{j-1}{r-1}}_{[\,r=j\,]}\hspace{14mm}f(r)\tag{2}\\
&=f(j)\tag{3}
\end{align}
$$
Explanation:
$(1)$: change order of summation and apply $\binom{a}{b}\binom{b}{c}=\binom{a}{c}\binom{a-c}{b-c}$
$(2)$: $\sum\limits_{t=r}^j(-1)^{t-r}\binom{j-r}{t-r}=\sum\limits_{t=r}^j\binom{j-r}{t-r}1^{j-t}(-1)^{t-r}=(1-1)^{j-r}$
$(3)$: the only non-zero term is $r=j$ and that term is $f(j)$
A: Note that it is:
$$
\begin{gathered}
  f(n) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)g(k)} \quad  \Leftrightarrow \quad g(n) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^{\,n - k} \left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)f(k)}  \hfill \\
  f(n) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)g(k)} \quad  \Leftrightarrow \quad g(n) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^{\,k} \left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)f(k)}  \hfill \\ 
\end{gathered} 
$$
because
$$
\begin{gathered}
  f(n) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)g(k)}  = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)\left( { - 1} \right)^{\,k - j} \left( \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right)f(j)} }  =  \hfill \\
   = \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)} {\left( {\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^{\,k - j} \left( \begin{gathered}
  n \\ 
  k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  k \\ 
  j \\ 
\end{gathered}  \right)} } \right)f(j)}  =  \hfill \\
   = \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n \\ 
  j \\ 
\end{gathered}  \right)\left( {\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( { - 1} \right)^{\,k - j} \left( \begin{gathered}
  n - j \\ 
  k - j \\ 
\end{gathered}  \right)} } \right)f(j)}  =  \hfill \\
   = \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n \\ 
  j \\ 
\end{gathered}  \right)\left( {\sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n - j \\ 
  k - j \\ 
\end{gathered}  \right)1^{\,n - j - \left( {k - j} \right)} \left( { - 1} \right)^{\,k - j} } } \right)f(j)}  =  \hfill \\
   = \sum\limits_{\left( {0\, \leqslant } \right)\,j\,\left( { \leqslant \,n} \right)} {\left( \begin{gathered}
  n \\ 
  j \\ 
\end{gathered}  \right)0^{\,n - j} f(j)}  = f(n) \hfill \\ 
\end{gathered} 
$$
where from 2nd to 3rd line we apply the known trinomial revision.  
In your case we have
$$
\begin{gathered}
  f(j) = \sum\limits_{\left( {1\, \leqslant } \right)\,t\,\left( { \leqslant \,j} \right)} {\sum\limits_{\left( {1\, \leqslant } \right)\,r\,\left( { \leqslant \,t} \right)} {\left( { - 1} \right)^{\,r + t} \left( \begin{gathered}
  j - 1 \\ 
  t - 1 \\ 
\end{gathered}  \right)\left( \begin{gathered}
  t - 1 \\ 
  r - 1 \\ 
\end{gathered}  \right)f(r)} }  =  \hfill \\
   = \sum\limits_{\left( {1\, \leqslant } \right)\,t\,\left( { \leqslant \,j} \right)} {\sum\limits_{\left( {1\, \leqslant } \right)\,r\,\left( { \leqslant \,t} \right)} {\left( { - 1} \right)^{\,r - t} \left( \begin{gathered}
  j - 1 \\ 
  t - 1 \\ 
\end{gathered}  \right)\left( \begin{gathered}
  t - 1 \\ 
  r - 1 \\ 
\end{gathered}  \right)f(r)} }  =  \hfill \\
   = \sum\limits_{\left( {1\, \leqslant } \right)\,t\,\left( { \leqslant \,j} \right)} {\sum\limits_{\left( {1\, \leqslant } \right)\,r\,\left( { \leqslant \,t} \right)} {\left( { - 1} \right)^{\,r - 1 - \left( {t - 1} \right)} \left( \begin{gathered}
  j - 1 \\ 
  t - 1 \\ 
\end{gathered}  \right)\left( \begin{gathered}
  t - 1 \\ 
  r - 1 \\ 
\end{gathered}  \right)f(r)} }  =  \hfill \\
   = \sum\limits_{\left( {1\, \leqslant } \right)\,t\,\left( { \leqslant \,j} \right)} {\sum\limits_{\left( {1\, \leqslant } \right)\,r\,\left( { \leqslant \,t} \right)} {\left( { - 1} \right)^{\,r - 1 - \left( {t - 1} \right)} \left( \begin{gathered}
  j - 1 \\ 
  t - 1 \\ 
\end{gathered}  \right)\left( \begin{gathered}
  t - 1 \\ 
  r - 1 \\ 
\end{gathered}  \right)g(r - 1)} }  =  \hfill \\
   = \sum\limits_{\left( {0\, \leqslant } \right)\,t - 1\,\left( { \leqslant \,j - 1} \right)} {\sum\limits_{\left( {0\, \leqslant } \right)\,r - 1\,\left( { \leqslant \,t - 1} \right)} {\left( { - 1} \right)^{\,r - 1 - \left( {t - 1} \right)} \left( \begin{gathered}
  j - 1 \\ 
  t - 1 \\ 
\end{gathered}  \right)\left( \begin{gathered}
  t - 1 \\ 
  r - 1 \\ 
\end{gathered}  \right)g(r - 1)} }  =  \hfill \\
   = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,j - 1} \right)} {\sum\limits_{\left( {0\, \leqslant } \right)\,l\,\left( { \leqslant \,k} \right)} {\left( { - 1} \right)^{\,l - k} \left( \begin{gathered}
  j - 1 \\ 
  k \\ 
\end{gathered}  \right)\left( \begin{gathered}
  k \\ 
  l \\ 
\end{gathered}  \right)g(l)} }  =  \hfill \\
   = g(j - 1) = f(j) \hfill \\ 
\end{gathered} 
$$
