Given $y_n=(1+\frac{1}{n})^{n+1}$ show that $\lbrace y_n \rbrace$ is a decreasing sequence
Note that $$ \begin{align} \frac{\left(1+\frac1n\right)^{n+1}}{\left(1+\frac1{n+1}\right)^{n+2}} &=\left(\frac{n+1}{n}\right)^{n+1}\left(\frac{n+1}{n+2}\right)^{n+2}\\ &=\frac{n}{n+1}\left(\frac{(n+1)^2}{n(n+2)}\right)^{n+2}\\ &=\frac{n}{n+1}\left(1+\frac1{n(n+2)}\right)^{n+2}\\ &\ge\frac{n}{n+1}\left(1+\frac{n+2}{n(n+2)}\right)\\ &=1 \end{align} $$ Therefore, $$ \left(1+\frac1{n+1}\right)^{n+2}\le\left(1+\frac1n\right)^{n+1} $$
Similarly, $$ \begin{align} \frac{\left(1+\frac1{n+1}\right)^{n+1}}{\left(1+\frac1n\right)^n} &=\left(\frac{n+2}{n+1}\right)^{n+1}\left(\frac{n}{n+1}\right)^n\\ &=\frac{n+1}{n}\left(\frac{n(n+2)}{(n+1)^2}\right)^{n+1}\\ &=\frac{n+1}{n}\left(1-\frac1{(n+1)^2}\right)^{n+1}\\ &\ge\frac{n+1}{n}\left(1-\frac{n+1}{(n+1)^2}\right)\\ &=1 \end{align} $$ Therefore, $$ \left(1+\frac1{n+1}\right)^{n+1}\ge\left(1+\frac1n\right)^n $$
Bernoulli's Inequality
In the preceding, we used Bernoulli's Inequality: for all $x\ge-1$ and non-negative integer $n$, $$ (1+x)^n\ge1+nx $$ This can be proven by induction:
Note that the inequality above is true for $n=0$.
Suppose that $x\ge-1$ and for a non-negative integer $n$, we have $$ (1+x)^n-nx\ge1 $$ Then $$ \begin{align} (1+x)^{n+1}-(n+1)x &=(1+x)^n-nx+x(1+x)^n-x\\ &\ge1+x((1+x)^n-1)\\ &\ge1 \end{align} $$ If $-1\le x\le0$, then both $x$ and $(1+x)^n-1$ are negative. If $x\ge0$, then both both $x$ and $(1+x)^n-1$ are positive. Therefore, if $x\ge-1$, $x((1+x)^n-1)\ge0$. This justifies the last inequality above.
Note that if $x\ne0$ and $n\ge1$, the last inequality is strict. Thus, for $x\ne0$ and $n\ge2$, we have $$ (1+x)^n\gt1+nx $$
Negative Exponents
Bernoulli's Inequality is also true for negative integer exponents. That is, for $x\gt-1$ and non-negative $n\in\mathbb{Z}$, $$ 1-nx\le(1+x)^{-n} $$ Suppose that $$ (1-nx)(1+x)^n\le1 $$ which is trivially true for $n=0$, and strictly true for $x\ne0$ and $n=1$. Then $$ \begin{align} (1-(n+1)x)(1+x)^{n+1} &=(1-nx)(1+x)^n-(n+1)x^2(1+x)^n\\ &\le1 \end{align} $$ Therefore, for all non-negative $n\in\mathbb{Z}$, $$ (1-nx)\le(1+x)^{-n} $$ where the inequality is strict for $x\ne0$ and $n\ge1$.
I think it needs just some basic algebraic manipulations: $$\frac{y_{n}}{y_{n+1}}=\frac{(1+\frac{1}n)^{n+1}}{(1+\frac{1}{n+1})^{n+2}}$$ You can show that the latter fraction is equal to $$(1+\frac{1}{n^2+2n})^{n+1}\times\frac{1}{1+1/(n+1)}$$ But $$(1+\frac{1}{n^2+2n})^{n+1}\ge {1+1/(n+1)}$$ Note that if $x\ge-1$ then $(1+x)^n\ge 1+nx.$
If you don't care about overkilling the Problem you can even do this with calculus, showing the derivative in n is negative. The derivative is $$ \frac{\left(\frac{1}{n}+1\right)^n (n+1) \left(n \log \left(\frac{1}{n}+1\right)-1\right)}{n^2}$$ and so we only need to show that $$1> n \log(1+\frac{1}{n})$$ By substitution $n=\frac{1}{x}$ we have $$\frac{\log(1+x)}{x}=\frac{\ln(1+x)-\ln(1)}{(x+1)-1}=\frac{1}{1+\xi}$$ with $\xi \in (0,x)$ (this is granted by the mean value theorem), and the last expression is less than 1.