1. **Stating the problem:** We are given inequalities involving logarithms and products, and we want to deduce the limit $$\lim_{n \to +\infty} \frac{\sqrt[n]{n!}}{n}$$. 2. **Recall the inequalities:** - For all $x > 0$, $$\ln(1+x) < x < (x+1)\ln(1+x)$$. - For integer $n \geq 1$, $$\prod_{k=1}^n \left(1+\frac{1}{k}\right)^k < e^n < \prod_{k=1}^n \left(1+\frac{1}{k}\right)^{k+1}$$. 3. **Rewrite the products:** Note that $$1 + \frac{1}{k} = \frac{k+1}{k}$$, so $$\prod_{k=1}^n \left(\frac{k+1}{k}\right)^k = \prod_{k=1}^n \frac{(k+1)^k}{k^k}$$. 4. **Simplify the product:** $$\prod_{k=1}^n \frac{(k+1)^k}{k^k} = \frac{2^1 3^2 4^3 \cdots (n+1)^n}{1^1 2^2 3^3 \cdots n^n}$$. 5. **Rewrite numerator and denominator:** The numerator is $$\prod_{j=2}^{n+1} j^{j-1}$$ and the denominator is $$\prod_{j=1}^n j^j$$. 6. **Express the product inequality:** $$\prod_{k=1}^n \left(1+\frac{1}{k}\right)^k = \frac{\prod_{j=2}^{n+1} j^{j-1}}{\prod_{j=1}^n j^j}$$. 7. **Use the inequality:** $$\prod_{k=1}^n \left(1+\frac{1}{k}\right)^k < e^n < \prod_{k=1}^n \left(1+\frac{1}{k}\right)^{k+1}$$. 8. **Relate to factorial and powers:** Using Stirling's approximation, $$n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$$, so $$\sqrt[n]{n!} \sim \frac{n}{e} \cdot (2\pi n)^{\frac{1}{2n}}$$. 9. **Evaluate the limit:** Since $(2\pi n)^{1/(2n)} \to 1$ as $n \to \infty$, $$\lim_{n \to +\infty} \frac{\sqrt[n]{n!}}{n} = \lim_{n \to +\infty} \frac{n/e \cdot (2\pi n)^{1/(2n)}}{n} = \frac{1}{e}$$. **Final answer:** $$\boxed{\lim_{n \to +\infty} \frac{\sqrt[n]{n!}}{n} = \frac{1}{e}}$$