A Familiar Limit?

$\begin{aligned} A & = \lim_{n \to \infty} \left(\frac {n^{n+1} + (n+1)^n}{n^{n+1}}\right)^n & \small \color{#3D99F6} \text{A }1^\infty \text{ case (see Reference)} \\ & = \exp \left( \lim_{n \to \infty} n\left(\frac {n^{n+1} + (n+1)^n}{n^{n+1}}-1 \right)\right) & \small \color{#3D99F6} \implies \lim_{n \to \infty} f(x)^{h(x)} = e^{\lim_{n \to \infty} h(x)(f(x)-1)} \\ & = \exp \left( \lim_{n \to \infty} \frac {n(n+1)^n}{n^{n+1}} \right) \\ & = \exp \left( \lim_{n \to \infty} \left(1+\frac 1n\right)^n \right) \\ & = e^e \approx 15.1534 \end{aligned}$

Therefore, $\lfloor 100A \rfloor = \boxed{1515}$ .

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Reference: $\color{#3D99F6} 1^\infty$ limit (2nd method)

$A=\lim_{n\to \infty}\left(\frac{n^{n+1}+(n+1)^{n}}{n^{n+1}}\right)^n$ $A=\lim_{n\to \infty}\left(1+\frac{(n+1)^{n}}{nn^{n}}\right)^n$ $A=\lim_{n\to \infty}\left(1+\frac{\left(\frac{n+1}{n}\right)^{n}}{n}\right)^n$ $A=\lim_{n\to \infty}\left(1+\frac{ \lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{n}}{n}\right)^n$ $A=\lim_{n\to \infty}\left(1+\frac{e}{n}\right)^n$ Making $p=\frac{n}{e}$ $A=\lim_{p\to \infty}\left(1+\frac{1}{p}\right)^{pe}$ $A=\left( \lim_{p\to \infty}\left(1+\frac{1}{p}\right)^p \right)^e$ $A=e^e$ $\lfloor{100A}\rfloor=1515$