If you meant what $\mathcal{O}((\log (n+7))^{\log (n)})$ is (so the power applies to the logarithm itself, not its argument), then we simplify that to $\mathcal{O}((\log (n))^{\log (n)})$; as per a suggestion by <PERSON>, here a proof for this equality:
Since $(\log (n))^{\log (n)}) \leq (\log (n+7))^{\log (n)})$ for all $n$, we have $\mathcal{O}((\log (n))^{\log (n)}) \subseteq \mathcal{O}((\log (n+7))^{\log (n)})$.
For the other direction, consider
$\qquad \begin{align} (\log (2n))^{\log (n)} &\leq c(\log (n))^{\log(n)} &\iff \ \frac{\log(2n)}{\log(n)} &\leq \sqrt[\log(n)]{c} &\iff \ 1+\frac{\log(2)}{\log(n)} &\leq \sqrt[\log(n)]{c} &\iff \ \left(1+\frac{\log(2)}{\log(n)}\right)^{\log(n)} &\leq c \end{align} $
What we now notice is that $\log n$ is a non-integer bound, so we can't directly expand to a polynomial here. We could if we knew that
$\left(1+\frac{\log(2)}{\log(n)}\right)^{\log(n)} \leq \left(1+\frac{\log(2)}{\lceil\log(n)\rceil}\right)^{\lceil\log(n)\rceil}$
Therefore, a small proof of the monotonicity of that function:
$\frac{d}{dx}\left(1+\frac{\log(2)}{\log(x)}\right)^{\log(x)} = \frac{\left(\frac{\log (2x)}{\log (x)}\right)^{\log (x)}\left(\log (2x)\log\left(\frac{\log (2x)}{\log (x)}\right) - \log (2)\right)}{x \log (2x)} \geq \frac{\log (2x)\log\left(\frac{\log (2x)}{\log (x)}\right) - \log (2)}{x (\log (x))^{\log (x)}} > 0$
for all $x > x_0$ for some $x_0$ [specifically, for $x_0 = 2$; check the link to math in the comment section].
Since we now know that the function is monotonically nondecreasing for all $n > 2$, we can choose our $c$ slightly differently:
\begin{align} \left(1+\frac{\log(2)}{\log(n)}\right)^{\log(n)} \leq \left(1+\frac{\log(2)}{\lceil\log(n)\rceil}\right)^{\lceil\log(n)\rceil} &\leq c &\overset{\text{rewriting the polynomial}}{\iff} \ \sum_{i=0}^{\lceil\log(n)\rceil}\binom{\lceil\log(n)\rceil}{i}\left(\frac{\log(2)}{\lceil\log(n)\rceil}\right)^i &\leq c \end{align}
Now we use the inequality $\binom{n}{k} \leq \frac{n^k}{k!}$ and see that \begin{align} \sum_{i=0}^{\lceil\log(n)\rceil}\binom{\lceil\log(n)\rceil}{i}\left(\frac{\log(2)}{\lceil\log(n)\rceil}\right)^i &\leq \sum_{i=0}^{\lceil\log(n)\rceil}\frac{(\lceil\log(n)\rceil)^i}{i! 532