Let $P_n(k,i)=|\{(d_1,\cdots,d_i): \ \sum_{j=1}^id_j=n, \ \forall j\ 1\leq d_j\leq k\}|$, the number of ordered partitions of $n$ into $i$ parts with individual parts bounded by $k$, with no piece having length 0. To be extra clear, if $n=9, \ k=4, \ i=3$ then $1+4+4=9$ and $4+1+4=9$ denote two different partitions.
I'm looking for elementary semi-decent upper bounds on $P_n(k,i)$ as $n$ gets large. We can assume $n\gg k$ and $[n/k]\leq i\leq n$.
For example $P_n(k,i)\leq k^{i-1}$, is the dumbest bound possible, combinatorially speaking. Another bound is $\binom{n}{i}$ by picking $i$ partitions. The second bound is somewhat worse but captures that $P_n(k,i)$ starts to get larger for $i>[n/k]$ but then ultimately goes down. Specifically we have $P_n(k,[t/k])=P(k,n)=1$.
Can one do significantly better than this without resorting to difficult estimates involving Gaussian polynomials, generating functions, etc.?
Generating functions are a powerful tool, you should have no fear in using them.
Here we have that $$P_n(k,i)=[z^n]\;(z+z^2+\ldots+z^k)^i = [z^n]\;z^i\left(\frac{1-z^k}{1-z}\right)^i = [z^{n-i}]\;\left(\frac{1-z^k}{1-z}\right)^i,$$ and: $$\frac{1}{(1-z)^i}=\sum_{m=0}^{+\infty}\binom{i+m-1}{m}z^m,\tag{1}$$ $$(1-z^k)^i = \sum_{r=0}^{i}\binom{i}{r}(-1)^r z^{kr},\tag{2}$$ so multiplying $(1)$ and $(2)$ we get: $$P_n(k,i)=\sum_{\substack{m+kr=n-i\\r\leq i}}\binom{i+m-1}{m}\binom{i}{r}(-1)^r\tag{3}.$$ Now the main contribute to this last sum is given by the first term $(r=0)$. The other ones are much smaller and, additionally, there is cancellation due to the alternating sign. So we have: