If you know how to deal with
$$n \int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx,$$
just split the integral at $\frac{\pi}{4}$ and see whether you can find something useful for the other part.
Since $x \mapsto \cos x - \sin x$ is strictly decreasing on $\bigl[0, \frac{\pi}{2}\bigr]$, and $\cos \frac{\pi}{4} = \sin \frac{\pi}{4}$, we have
$$\lvert \cos x - \sin x\rvert \leqslant c := \sin 1 - \cos 1$$
for $\frac{\pi}{4} \leqslant x \leqslant 1$. Hence
$$\Biggl\lvert \int_{\frac{\pi}{4}}^1 (\cos x - \sin x)^n\,dx\Biggr\rvert \leqslant \int_{\frac{\pi}{4}}^1 c^n\,dx < c^n.$$
Since $0 < c < 1$, it follows that
$$n \int_{\frac{\pi}{4}}^1 (\cos x - \sin x)^n\,dx \to 0.$$
That is useful indeed. Thus we need only consider
$$\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx = 2^{n/2} \int_0^{\frac{\pi}{4}} \sin^n \biggl(\frac{\pi}{4} - x\biggr)\,dx = 2^{n/2}\int_0^{\frac{\pi}{4}} \sin^n x\,dx.$$
Substituting $u = \sin x$, then $v = \sqrt{2}\cdot u$, and afterwards integrating by parts, we find (for $n \geqslant 1$)
\begin{align}
n\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx
&= n 2^{n/2} \int_0^{\frac{\pi}{4}} \sin^n x\,dx \\
&= n 2^{n/2} \int_0^{\frac{1}{\sqrt{2}}} \frac{u^n}{\sqrt{1-u^2}}\,du \\
&= n \int_0^1 \frac{v^n}{\sqrt{2-v^2}}\,dv \\
&= v^n\cdot \frac{v}{\sqrt{2-v^2}}\biggr\rvert_0^1 - \int_0^1 v^n\biggl(\frac{1}{\sqrt{2-v^2}} + \frac{v^2}{(2-v^2)^{3/2}}\biggr)\,dv \\
&= 1 - \int_0^1 \frac{2v^n}{(2-v^2)^{3/2}}\,dv.
\end{align}
Since $\frac{1}{\sqrt{2}} < \frac{2}{(2-v^2)^{3/2}} < 2$ for $0 < v < 1$, we thus have
$$1 - \frac{2}{n+1} < n\int_0^{\frac{\pi}{4}} (\cos x - \sin x)^n\,dx < 1 - \frac{1}{\sqrt{2}\,(n+1)}.$$