"Unrolling" a recurrence relation
Let $S(n) = T(n)/(n+1)$. Notice that $S(0) = 0$ and $S(n) = S(n-1) + 1/n(n-1)$. Thus
$$
S(n) = \frac{1}{n(n+1)} + S(n-1) = \frac{1}{n(n+1)} + \frac{1}{(n-1)n} + S(n-2) = \cdots
$$
Continuing in this way, for any $i \geq 0$ we get
$$
S(n) = \frac{1}{n(n+1)} + \frac{1}{(n-1)n} + \cdots + \frac{1}{(i+1)(i+2)} + S(i).
$$
In particular, for $i = 0$ we get
$$
S(n) = \frac{1}{n(n+1)} + \frac{1}{(n-1)n} + \cdots + \frac{1}{(1)(2)}.
$$



Alternatively, you can prove by induction that $S(n) = n/(n+1)$. This clearly holds for $n = 0$. Assuming it holds for $n-1$, we prove it for $n$:
$$
S(n) = \frac{1}{n(n+1)} + S(n-1) = \frac{1}{n(n+1)} + \frac{n-1}{n} = \frac{1+(n-1)(n+1)}{n(n+1)} = \frac{n^2}{n(n+1)} = \frac{n}{n+1}.
$$



Alternatively, we can prove directly that $T(n) = n$ by induction. This clearly holds for $n = 0$. Assuming it holds for $n-1$, we prove it for $n$:
$$
T(n) = \frac{n+1}{n} T(n-1) + \frac{n+1}{n(n+1)} = \frac{n+1}{n} \cdot (n-1) + \frac{1}{n} = \frac{(n+1)(n-1)+1}{n} = \frac{n^2}{n} = n.
$$