How prove $\frac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{n}\frac{1}{t^2}<e$

Let $k,n\in \mathbb{N},n\ge k$, prove that $$\dfrac{(k+1)^{k+1}}{k^k}\sum_{t=k+1}^{n}\dfrac{1}{t^2}<e.$$

I got the impression that this inequality is very sharp.

My idea: $$\sum_{t=k+1}^{n}\dfrac{1}{t^2}\le\sum_{t=k+1}^{n}\dfrac{1}{t(t-1)}=\dfrac{1}{k}-\dfrac{1}{n},$$

$$\Longleftrightarrow \dfrac{(k+1)^{k+1}}{k^k}\left(\dfrac{1}{k}-\dfrac{1}{n}\right)<\dfrac{(k+1)^{k+1}}{k^k}\dfrac{1}{k}<e$$ $$\Longleftrightarrow \left(1+\dfrac{1}{k}\right)^{k+1}<e.$$ It is well konwn that $$(1+1/x)^{x+1}>e,x>0.$$ so I failed at this direction. Thank you everyone help.

Idea 2: use the well-known $$\left(1+\dfrac{1}{x}\right)^x<e\left(1-\dfrac{1}{2(1+x)}\right)$$ $$\Longleftrightarrow \dfrac{n-k}{k+1}\left(1-\dfrac{1}{2(k+1)}\right)<1,$$ but this can not prove the inequality either.

A sketch. First note that:

$$\sum_{r=m+1}^{\infty} \frac{1}{r^2}=\int_0^1 \frac{x^m \ln x}{x-1}\,dx$$

Now:

$$\begin{aligned}\left(1+\frac{1}{k}\right)^{k+1} \sum_{t=k+1}^n \frac{k}{t^2}&<\left(1+\frac{1}{k}\right)^{k+1} \sum_{t=k+1}^{\infty} \frac{k}{t^2}\\&=\left(1+\frac{1}{k}\right)^{k+1}\int_0^1 \frac{kx^k \ln x}{x-1}\,dx\\&<\left(1+\frac{1}{k}\right)^{k+1}\int_0^1 kx^{k-\frac{1}{2}} \,dx\\&=\left(1+\frac{1}{k}\right)^{k+1}\left(\frac{2k}{2k+1}\right)\\&=\left(1+\frac{1}{k}\right)^k\frac{2k+2}{2k+1}\\&<e\end{aligned}$$