Let $t$ be a fixed positive integer. I am interested in counting the number of tuples of positive integers that sum to $t$. I don't care about the size of the tuples, so the $1$-tuple $(t)$ is valid, as is the $t$-tuple $(1,1,...,1)$.
I start by fixing the length of the tuple at $k$, $1\leq k \leq t$ and then summing over $k$. The problem is I get stuck here. This is similar to the problem of dividing up $t$ dots by using $k-1$ lines, which gives $$\binom{t+k-1}{ k-1}$$, but this would also count non-negative solutions but I'm looking for strictly positive ones, since here the solution $(0, t-1, 1)$ is the same as the solution $(t-1,1)$. So how can I alter this to count only solutions which are strictly positive?
What you're after is called the number of compositions of $t$. The answer is known to be $2^{t-1}$.
The proof given on that Wikipedia article goes as follows: create a tuple by starting with $t$ $1$s in a row with boxes between them: $$ (1\square 1\square\cdots\square 1) $$ Now for each box, decide whether to put a $+$ or a $,$ in it. This can be done in $2^{t-1}$ ways, each way gives exactly one composition, and each composition may be constructed in exactly one way.
If you want to continue your proof, first assign $1$ to each entry in the tuple, then use your method to distribute the remaining $t-k$ among the entries. This gives $\binom{t-1}{k-1}$ tuples of length $k$ (this corresponds to distributing $k-1$ commas in the above proof). Summing all these up for $k$ from $1$ to $t$ will give you $2^{t-1}$.