= Solution
Write $\vartheta(x)=\sum_{p\le x}\log p$ for the <Chebyshev theta function> and $\psi(x)=\sum_{p^j\le x}\log p$ for the <Second Chebyshev function>. Throughout, $p$ denotes a <prime> and all <logarithms> are natural. Each <prime> $n<p\le2n$ divides the central <binomial coefficient>, so
$$
\vartheta(2n)-\vartheta(n)\le\log\binom{2n}{n}\le2n\log2.
$$
Apply this at $n=1,2,4,\ldots,2^{r-1}$ and telescope: $\vartheta(2^r)\le(2^{r+1}-2)\log2$. Bounding a general $x$ by the next power of two proves $\vartheta(x)=O(x)$. Splitting the <prime-counting function> at $\sqrt x$ now gives
$$
\pi(x)\le\sqrt x+\frac2{\log x}\vartheta(x),
\qquad\boxed{\pi(x)=O(x/\log x).}
$$
This is the required <Chebyshev estimate from central binomial coefficients>, with no use of a <prime> asymptotic.
The higher <prime powers> also satisfy
$$
\psi(x)=\sum_{1\le j\le\log x/\log2}\vartheta(x^{1/j})
=\vartheta(x)+O(\sqrt x\log x)=O(x).
$$
For an <integer> $N$, the <prime factorization> of a <factorial> gives
$$
\log N!=\sum_{p^j\le N}\left\lfloor\frac N{p^j}\right\rfloor\log p
=N\sum_{p^j\le N}\frac{\log p}{p^j}+O(\psi(N)).
$$
The <Stirling formula>, or integral comparison for $\log t$, says $\log N!=N\log N-N+O(\log N)$. After division by $N$ we obtain $\sum_{p^j\le N}(\log p)/p^j=\log N+O(1)$. Moreover
$$
0\le\sum_{p}\sum_{j\ge2}\frac{\log p}{p^j}
=\sum_p\frac{\log p}{p(p-1)}
\le\sum_{n\ge2}\frac{\log n}{n(n-1)}<\infty.
$$
Subtract this bounded contribution and replace $N$ by $\lfloor x\rfloor$. The <factorial proof of Mertens first theorem> therefore yields
$$
\boxed{A(x):=\sum_{p\le x}\frac{\log p}{p}=\log x+O(1).}
$$
For fixed $\lambda>1$, apply <partial summation> to $A(t)$ and $f(t)=(\log t)^{\lambda-1}$:
$$
\sum_{p\le x}\frac{(\log p)^\lambda}{p}
=A(x)(\log x)^{\lambda-1}
-(\lambda-1)\int_2^x A(t)\frac{(\log t)^{\lambda-2}}t\,dt.
$$
The main contribution is
$$
(\log x)^\lambda-\frac{\lambda-1}{\lambda}(\log x)^\lambda+O(1)
=\frac1\lambda(\log x)^\lambda+O(1).
$$
The bounded error in $A(t)$ contributes $O((\log x)^{\lambda-1})$ at the endpoint and in the integral, since $\int_2^x(\log t)^{\lambda-2}dt/t=O_\lambda((\log x)^{\lambda-1})$. Thus the <positive logarithmic moments of reciprocal primes> satisfy
$$
\boxed{\sum_{p\le x}\frac{(\log p)^\lambda}{p}
=\frac1\lambda(\log x)^\lambda+O_\lambda((\log x)^{\lambda-1}).}
$$
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