Partial sum of binomial coefficents I am attempting to show that:
$$
\sum\limits_{q=0}^Q\binom{q + d - 1}{d - 1} = (Q + 1)d^Q
$$
But I am totally lost. I was hoping I could use something as simple as the binomial theorem, but I quickly get stuck. Any help is greatly appreciated...
EDIT:
Thanks to @stochasticboy321 for your answer! I see now that the equality cant hold... Im now trying to prove by induction a weaker, but as useful claim for my purpose, that:
$$
\sum\limits_{q=0}^Q\binom{q + d - 1}{d - 1} \leq (Q + 1)d^Q
$$
for $d > 0$ and $Q > 0$. But Ive gotten stuck once more :D.
Base case, $Q = 1$:
$$
\binom{d - 1}{d - 1} + \binom{1 + d - 1}{d - 1} \leq 2d \implies 1 + d \leq 2d
$$
Assume $\sum_{q=0}^k\binom{q + d - 1}{d - 1} \leq (k + 1)d^k$, prove:
$$
\sum\limits_{q=0}^{k+1}\binom{q + d - 1}{d - 1} \leq (k + 2)d^{k + 1} \\
\implies \sum\limits_{q=0}^k\binom{q + d - 1}{d - 1} + \binom{k + d}{d - 1} \leq (k + 2)d^{k + 1} \\
\implies (k + 1)d^k + \binom{k + d}{d - 1} \leq (k + 2)d^{k+1}
$$
My strategy so far has been trying to expand $\binom{k + d}{d - 1}$ to its factorial form and reduce, but I really dont know what Im doing...
 A: As noted in the comments, the identity asked is incorrect, and the correct identity is $$\sum\limits_{q=0}^Q\binom{q + d - 1}{d - 1} = \frac{Q+1}{d} \binom{Q+d}{d-1} = \binom{Q+d}{d}.$$ The following is a combinatorial proof of the same. Note that, by stars and bars, $\binom{q+d-1}{d-1}$ is the number of non-negative integer solutions to the equation $x_1+x_2+ \dots + x_d = q$. Thus, the summation is the number of non-negative integer solutions to $x_1+x_2+\dots + x_d \le Q$. But each non-negative integer solution of $x_1+x_2+\dots + x_d \le Q$ can be identified with a non-negative integer solution of $x_1+x_2+\dots + x_d +s = Q$, and the latter has $\binom{Q+d}{d}$ solutions. QED.

Regarding you edit - it's much easier to go inductively in $d$. 
Claim: For $Q,d \ge 1, \binom{Q+d}{d} \le (Q+1)d^Q.$
Pf. Note that for every $Q$, $\binom{Q+0}{0} = Q \le Q+1,$  and thus the result holds for $d = 1$. We assume the result for $(Q,d)$. Now, for $Q,d \ge 1$,\begin{align*}
(d+1)^{Q+1} \overset{(a)}\ge& d^{Q+1} + (Q+1)d^Q = d^Q(Q+d+1)\\
\iff (Q+1)(d+1)^Q \ge& \frac{(Q+1)d^Q (Q+d+1)}{d+1} \\
\overset{(b)}\implies (Q+1)(d+1)^Q \ge& \binom{Q+d}{d} \frac{Q+d+1}{d+1} = \binom{Q+d+1}{d+1}
\end{align*}
where inequality $(a)$ is because each term in the binomial theorem expansion of $(d+1)^{Q+1}$ is positive, and implication $(b)$ is  due to the induction hypothesis. QED.
A: Algebraic Proof
Using formulas from GouldBk.pdf    http://www.dsi.dsi.unifi.it/~resp/GouldBK.pdf
Partial sum ((P) in GouldBk.pdf page 14)
${\displaystyle \sum_{k=0}^{n}f_{k}=\left[t^{n}\right]\frac{f(t)}{1-t}}$
Let's rewrite the sum as
${\displaystyle \sum_{q=0}^{Q}\left(\begin{array}{c}
q+d-1\\
q
\end{array}\right)}$
and apply (1.49 on page 49)
${\displaystyle =\left(\begin{array}{c}
d+Q\\
Q
\end{array}\right)}$
Which matches the previous solution.
I couldn't do better than the 1.49 proof, which uses (BC4)
Proof (directly lifted):
${\displaystyle \sum_{q=0}^{Q}\left(\begin{array}{c}
d-1+q\\
q
\end{array}\right)=\left[t^{Q}\right]\frac{1}{\left(1-t\right)^{\left(d-1+1\right)}\left(1-t\right)}}=\left[t^{Q}\right]\frac{1}{\left(1-t\right)^{\left(d+1\right)}}=\left(\begin{array}{c}
d+Q\\
Q
\end{array}\right)$
