If $$n(2n + 1) \displaystyle\int_0^1 (1 - x^n)^{2n} dx = 1177 \int_0^1 (1 - x^n)^{2n+1} dx$$, $$n \in \mathbb{N}$$, then $$n$$ is equal to ______.

We need to find $$n \in \mathbb{N}$$ such that $$n(2n+1)\int_0^1 (1-x^n)^{2n}\,dx = 1177\int_0^1 (1-x^n)^{2n+1}\,dx$$.

Let $$I_m = \int_0^1 (1 - x^n)^m\,dx$$.

Substituting $$u = x^n\,,\; x = u^{1/n}\,,\; dx = \frac{1}{n}u^{1/n - 1}\,du$$ transforms $$I_m$$ into $$I_m = \frac{1}{n}\int_0^1 u^{1/n - 1}(1-u)^m\,du$$.

Recall the Beta function: $$B(a,b)=\int_0^1 u^{a-1}(1-u)^{b-1}\,du=\frac{\Gamma(a)\,\Gamma(b)}{\Gamma(a+b)}$$.

Here $$a=\frac{1}{n}$$ and $$b=m+1$$, so $$I_m=\frac{1}{n}\,B\Bigl(\frac{1}{n},\,m+1\Bigr)=\frac{1}{n}\,\frac{\Gamma(1/n)\,\Gamma(m+1)}{\Gamma(m+1+1/n)}$$.

Next, using this formula gives $$\frac{I_{2n}}{I_{2n+1}} = \frac{\Gamma(2n+1)\,\Gamma(2n+2+1/n)}{\Gamma(2n+2)\,\Gamma(2n+1+1/n)}$$.

Using the Gamma function property $$\Gamma(s+1)=s\,\Gamma(s)$$ yields $$\frac{\Gamma(2n+1)}{\Gamma(2n+2)}=\frac{1}{2n+1}$$ and $$\frac{\Gamma(2n+2+1/n)}{\Gamma(2n+1+1/n)}=2n+1+\frac{1}{n}$$.

Thus $$\frac{I_{2n}}{I_{2n+1}}=\frac{2n+1+1/n}{2n+1}=1+\frac{1}{n(2n+1)}=\frac{n(2n+1)+1}{n(2n+1)}$$.

The equation $$n(2n+1)\,I_{2n}=1177\,I_{2n+1}$$ gives $$\frac{I_{2n}}{I_{2n+1}}=\frac{1177}{n(2n+1)}$$.

Setting this equal to the previous result leads to $$\frac{n(2n+1)+1}{n(2n+1)}=\frac{1177}{n(2n+1)}$$, so $$n(2n+1)+1=1177$$ and hence $$n(2n+1)=1176$$, so $$2n^2+n-1176=0$$.

Next, the quadratic formula gives $$n=\frac{-1\pm\sqrt{1+4\times2\times1176}}{4}=\frac{-1\pm\sqrt{9409}}{4}$$, and since $$\sqrt{9409}=97$$ we have $$n=\frac{-1+97}{4}=\frac{96}{4}=24$$ (the negative root is rejected).

The correct answer is $$\boxed{24}$$.