The following common inequalities suffice here:
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\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$
$1+x+\lfrac12 x^2 \le \exp(x) \le 1 + x + x^2$ for every real $x \in [0,\ln(2)]$.
$1 - \lfrac12 x^2 \le \cos(x) \le 1 - \lfrac12 x^2 + \lfrac1{24} x^4$ for every real $x$.
$x - \lfrac16 x^3 \le \sin(x) \le x$ for every real $x \ge 0$.
They can be proven elementarily by recursively comparing derivatives.
As $n \to \infty$:
Let $r = n^{2/3}$.
Given any $x \in [0,\lfrac1r]$:
$\cos(x) - \sin(x) \le 1 - x + \lfrac16 x^3 \le \exp(-x)$.
$\cos(x) - \sin(x) \ge 1 - x - \lfrac12 x^2 \ge \exp(-x-2x^2)$ since $x \to 0$.
Thus $(\cos(x)-\sin(x))^n \in \exp(-x) ^ n \cdot [\exp(-\frac{2}{r^2}),1]^n$.
Thus ${\displaystyle\int}_0^{\lfrac1r} n ( \cos(x) - \sin(x) )^n\ dx \in [\exp(-\frac{2n}{r^2}),1] \cdot {\displaystyle\int}_0^{\lfrac1r} n \exp(-nx)\ dx$.
Also ${\displaystyle\int}_0^{\lfrac1r} n \exp(-nx)\ dx = 1 - \exp(-\lfrac{n}{r}) \to 1$.
Given any $x \in [\lfrac1r,1]$:
$\cos(x) - \sin(x) \le 1 - x - x^2 ( \lfrac12 - \lfrac16 x - \lfrac1{24} x^2 ) \le 1 - \lfrac1r$.
$\cos(x) - \sin(x) \ge 1 - x - \lfrac12 x^2 \ge -\frac12$.
Thus $| n ( \cos(x) - \sin(x) )^n | \le n ( 1 - \lfrac1r )^n \to 0$.
Thus ${\displaystyle\int}_{\lfrac1r}^1 n ( \cos(x) - \sin(x) )^n\ dx \to 0$.
Therefore ${\displaystyle\int}_0^1 n ( \cos(x) - \sin(x) )^n\ dx \to 1$.