combinatorics - 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 $) ...

Answer

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

$$\sum_{q=1}^n f(q) {n-1\choose q-1} [[n-q = 0]]
= f(n) {n-1\choose n-1} = f(n).$$