I'm trying to show that for a distribution with a finite support with probabilities $p_1,\ldots,p_n$, we get that with $k\rightarrow\infty$ the probability $p^k$ becomes uniform over ${k}\choose{p_1k,\ldots,p_nk}$ size support.
My final goal is to define another distribution over this support, and to show that the $L_1$ norm distance between them over the $p^k$ support approaches zero.
I tried to use the law of big numbers, but either I use the almost surely convergence, which implies that the event $P(\lim_{k\rightarrow\infty}\frac{1}{k}\{\text{number of times }p_i\text{ happened}\}=p_i)=1 $ but gives me no direct information of the probabilities of the events which aren't in the ${k}\choose{p_1k,\ldots,p_nk}$ subspace, for which I wish to find a boundary that approaches $0$ as $k\rightarrow\infty$, neither tells me that $\lim_{k\rightarrow\infty}P(\frac{1}{k}\{\text{number of times }p_i\text{ happened}\}=p_i)=1$, or I use the $P$ convergence which tells me $\forall \varepsilon>0 \quad \lim_{k\rightarrow\infty} P(\vert\{\text{number of times }p_i\text{ happened}\}-p_ik\vert > k\varepsilon)=0$ but these constant "margins" make it impossible for me to say that as $k\rightarrow\infty$, the sum of probabilities of events outside of the sub-support ${k}\choose{p_1k,\ldots,p_nk}$ vanishes.
Actually, I believe it's not quite true, as I tried a different approach: They probability of each event in the ${k}\choose{p_1k,\ldots,p_nk}$ subspace is $\prod_{i=1}^n p_i^{p_ik}$, thus the probability of an event in this subspace is: $${k \choose p_{1}k,p_{2}k,\ldots,p_{s}k}\prod_{i=1}^{s}p_{i}^{p_{i}k}=\frac{k!}{\prod_{i=1}^{s}kp_{i}!}\prod_{i=1}^{s}p_{i}^{p_{i}k} \\ \frac{k!}{\prod_{i=1}^{s}kp_{i}!}\prod_{i=1}^{s}p_{i}^{p_{i}k}\sim\frac{\sqrt{2\pi k}\left(\frac{k}{e}\right)^{k}}{\prod_{i=1}^{s}\sqrt{2\pi kp_{i}}\left(\frac{kp_{i}}{e}\right)^{kp_{i}}}\prod_{i=1}^{s}p_{i}^{p_{i}k} \\ =\frac{\sqrt{2\pi k}\left(\frac{k}{e}\right)^{k}}{\prod_{i=1}^{s}\sqrt{2\pi kp_{i}}\left(\frac{k}{e}\right)^{kp_{i}}} \\ =\frac{\sqrt{2\pi k}\left(\frac{k}{e}\right)^{k}}{\left(\frac{k}{e}\right)^{\underbrace{\left(\sum_{i}p_{i}\right)}_{=1}k}\prod_{i=1}^{s}\sqrt{2\pi kp_{i}}} \\ =\frac{\sqrt{2\pi k}}{\prod_{i=1}^{s}\sqrt{2\pi kp_{i}}}\rightarrow0,\quad k\rightarrow\infty$$
However, I know that the $p^k$ distribution does approach a uniform distribution over this subspace as $k\rightarrow\infty$.
Where have a I gone wrong, and how can I prove that the probabilities outside this sub-support vanish?