Prove that for a sequence $(n_k)_{k\in\mathbb{N}}$ in $\mathbb{N}$, with $(n_k)$ strictly monotone increasing, $n_k\geqslant k$, for all $k\in\mathbb{N}$.
I started with noticing that $n_0\geqslant 0$ and $n_1>n_0$, so $n_1\geqslant1$. So we see that $n_k\geqslant k$ for $k=1$ and $k=2$. Now, suppose that $n_j\geqslant j$ for $j=\{0,1,\dotsc,k\}$. We see that $n_{k+1}>n_k\geqslant k$.
But from here I'm stuck. Does anyone have something to keep me going?
Since $n_{k+1}>n_k$ and $n_k\ge k\implies n_{k+1}>k\implies n_{k+1}\ge k+1$
DETAILS
Note that $(n_k)$ is strictly increasing so $n_{k+1}> n_k$
And since $n_k\ge k$
So we must have $n_{k+1}>n_k\ge k$
Now since $n_k\in \Bbb N$ for each $n\in \Bbb N$ and $n_{k+1}>k$ we must have $n_{k+1}\ge k+1$