How to solve a recurrence relation with a sum?
Here are several ways to solve your recurrence relation.

Guessing

Anyone with enough experience in computer science might recognize your recurrence as the one satisfied by $T(n) = 2^n$. Given this guess, you can verify it by summing the appropriate geometric series: if $T(m) = 2^m$ for $m &lt; n$ then

$$
T(n) = 1 + \sum_{m=0}^{n-1} T(m) = 1 + \sum_{m=0}^{n-1} 2^m = 1 + (2^n-1) = 2^n.
$$

Computing the first few values

Another approach which is very  useful is to compute a few values of the sequence:
$$ 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, ... $$
If we're lucky, as in this case, we might spot a pattern. If not, it's always a good idea to consult the on-line encyclopedia of integer sequences. A quick search would reveal the sequence A000079.

Creative telescoping

In the method of creative telescoping, we try to find a combination of terms that results in major cancellation. In our case, we can try $T(n+1)-T(n)$:

$$
T(n+1)-T(n) = \sum_{m=0}^n T(m) - \sum_{m=0}^{n-1} T(m) = T(n),
$$
and so $T(n+1) = 2T(n)$. Unrolling this rule, we deduce $T(n) = 2^n T(0) = 2^n$.

Back-substitution

Another way to look at the preceding trick is to try to "back-substitute" the recurrence inside itself:

$$
T(n) = 1 + \sum_{m=0}^{n-1} T(m) = 1 + \sum_{m=0}^{n-2} T(m) + T(n-1) = 2T(n-1),
$$

from which we can proceed as before.

Generating series

A very general method is that of generating series. Let $G(x) = \sum_{n=0}^\infty T(n) x^n$. Since $1/(1-x) = \sum_{n=0}^\infty x^i$, multiplying by $1/(1-x)$ is like computing "running sums":
$$
\frac{G(x)}{1-x} =
\sum_{n=0}^\infty \sum_{m=0}^n T(m) x^n =
\sum_{n=0}^\infty (T(n+1)-1) x^n = \\ \frac{G(x)-T(0)}{x} - \sum_{n=0}^\infty x^n = \frac{G(x)-1}{x} - \frac{1}{1-x}.
$$
From this it is not hard to deduce
$$
G(x) = \frac{-1/x-1/(1-x)}{1/(1-x)-1/x} = \frac{-(1-x)-x}{x-(1-x)} = \frac{1}{1-2x} = \sum_{n=0}^\infty 2^nx^n,
$$
and so $T(n) = 2^n$.

Recursion tree

It is also possible to deduce the value of $T(n)$ using a recursion tree. Each path in the recursion tree for $T(n)$ corresponds to a sequence of decreasing numbers $n_0 = n, \ldots, n_\ell$, where $n_\ell \geq 0$ (possibly $\ell = 0$); this is because all leaves are labeled by $1$. The set of possible decreasing sequences is in one-to-one correspondence with subsets of $\{n-1,\ldots,0\}$, because once we know which numbers appear, we also know their order. The number of subsets of $\{n-1,\ldots,0\}$ is exactly $2^n$.